Class 12-NCERT Solutions-Chapter-07-Integrals-Mis

Solution:

\[ \frac{1}{x-x^3} = \frac{1}{x(1-x^2)} = \frac{1}{x(1-x)(1+x)} \]

Let \(\frac{1}{x(1-x)(1+x)} = \frac{A}{x} + \frac{B}{1-x} + \frac{C}{1+x} \quad \dots(1)\)

\(\Rightarrow 1 = A(1-x^2) + Bx(1+x) + Cx(1-x)\)
\(\Rightarrow 1 = A - Ax^2 + Bx + Bx^2 + Cx - Cx^2\)

Equating the coefficients of \(x^2\), \(x\), and constant term, we obtain:

\(-A + B - C = 0\)
\(B + C = 0\)
\(A = 1\)

On solving these equations, we obtain:

\(A = 1, B = \frac{1}{2}, \text{ and } C = -\frac{1}{2}\)

From equation (1), we obtain:

\[ \frac{1}{x(1-x)(1+x)} = \frac{1}{x} + \frac{1}{2(1-x)} - \frac{1}{2(1+x)} \]

Integrating both sides with respect to \(x\):

\[ \Rightarrow \int \frac{1}{x(1-x)(1+x)}dx = \int \frac{1}{x}dx + \frac{1}{2} \int \frac{1}{1-x}dx - \frac{1}{2} \int \frac{1}{1+x}dx \]

\[ = \log|x| - \frac{1}{2} \log|1-x| - \frac{1}{2} \log|1+x| \] \[ = \log|x| - \log\left|(1-x)^{\frac{1}{2}}\right| - \log\left|(1+x)^{\frac{1}{2}}\right| \] \[ = \log \left| \frac{x}{(1-x)^{\frac{1}{2}}(1+x)^{\frac{1}{2}}} \right| + C \] \[ = \log \left| \left( \frac{x^2}{1-x^2} \right)^{\frac{1}{2}} \right| + C \] \[ = \frac{1}{2} \log \left| \frac{x^2}{1-x^2} \right| + C \]

Solution:

First, we rationalize the denominator by multiplying and dividing by the conjugate:

\[ \begin{aligned} \frac{1}{\sqrt{x+a} + \sqrt{x+b}} &= \frac{1}{\sqrt{x+a} + \sqrt{x+b}} \times \frac{\sqrt{x+a} - \sqrt{x+b}}{\sqrt{x+a} - \sqrt{x+b}} \\ &= \frac{\sqrt{x+a} - \sqrt{x+b}}{(x+a) - (x+b)} \\ &= \frac{\sqrt{x+a} - \sqrt{x+b}}{a-b} \end{aligned} \]

Now, integrating the function:

\[ \Rightarrow \int \frac{1}{\sqrt{x+a} + \sqrt{x+b}} dx = \frac{1}{a-b} \int \left( \sqrt{x+a} - \sqrt{x+b} \right) dx \]

Using the power rule \(\int x^n dx = \frac{x^{n+1}}{n+1}\):

\[ = \frac{1}{(a-b)} \left[ \frac{(x+a)^{\frac{3}{2}}}{\frac{3}{2}} - \frac{(x+b)^{\frac{3}{2}}}{\frac{3}{2}} \right] \] \[ = \frac{2}{3(a-b)} \left[ (x+a)^{\frac{3}{2}} - (x+b)^{\frac{3}{2}} \right] + C \]

Solution:

Let \(x = \frac{a}{t} \Rightarrow dx = -\frac{a}{t^2}dt\). Substituting these values into the integral:

\[ \begin{aligned} \Rightarrow \int \frac{1}{x\sqrt{ax-x^2}}dx &= \int \frac{1}{\frac{a}{t}\sqrt{a\cdot\frac{a}{t} - \left(\frac{a}{t}\right)^2}} \left(-\frac{a}{t^2}dt\right) \\ &= -\int \frac{1}{at} \cdot \frac{1}{\sqrt{\frac{a^2}{t} - \frac{a^2}{t^2}}} dt \\ &= -\frac{1}{a} \int \frac{1}{\sqrt{\frac{t^2}{t} - \frac{t^2}{t^2}}} dt \\ &= -\frac{1}{a} \int \frac{1}{\sqrt{t-1}} dt \end{aligned} \]

Integrating with respect to \(t\):

\[ = -\frac{1}{a} [2\sqrt{t-1}] + C \]

Substituting \(t = \frac{a}{x}\) back:

\[ = -\frac{1}{a} \left[ 2\sqrt{\frac{a}{x}-1} \right] + C \] \[ = -\frac{2}{a} \sqrt{\frac{a-x}{x}} + C \]

Solution:

Multiplying and dividing the numerator and denominator by \(x^{-3}\) (or effectively factoring \(x^4\) out of the bracket):

\[ \frac{1}{x^2(x^4+1)^{\frac{3}{4}}} = \frac{1}{x^2 \cdot (x^4)^{\frac{3}{4}} \left(1+\frac{1}{x^4}\right)^{\frac{3}{4}}} = \frac{1}{x^5 \left(1+\frac{1}{x^4}\right)^{\frac{3}{4}}} \]

Let \(1 + \frac{1}{x^4} = t\). Differentiating gives \(-\frac{4}{x^5} dx = dt \Rightarrow \frac{1}{x^5} dx = -\frac{dt}{4}\).

Substituting into the integral:

\[ \int \frac{1}{x^5 \left(1+\frac{1}{x^4}\right)^{\frac{3}{4}}} dx = -\frac{1}{4} \int \frac{1}{t^{\frac{3}{4}}} dt \] \[ = -\frac{1}{4} \int (1+t)^{-\frac{3}{4}} dt \] \[ = -\frac{1}{4} \left[ \frac{t^{\frac{1}{4}}}{\frac{1}{4}} \right] + C \] \[ = - (t)^{\frac{1}{4}} + C \]

Substituting back \(t = 1 + \frac{1}{x^4}\):

\[ = -\left(1+\frac{1}{x^4}\right)^{\frac{1}{4}} + C \]

Solution:

\[ \frac{1}{x^{1/2} + x^{1/3}} = \frac{1}{x^{1/3}(1 + x^{1/6})} \]

Let \(x = t^6 \Rightarrow dx = 6t^5 dt\).

Substituting this into the integral:

\[ \Rightarrow \int \frac{1}{x^{1/2} + x^{1/3}} dx = \int \frac{6t^5}{t^3(1+t)} dt = 6 \int \frac{t^2}{1+t} dt \]

Performing polynomial long division or algebraic manipulation (\(t^2 = t^2 - 1 + 1\)):

\[ = 6 \int \left( (t-1) + \frac{1}{1+t} \right) dt \] \[ = 6 \left[ \frac{t^2}{2} - t + \log|1+t| \right] \] \[ = 3t^2 - 6t + 6\log|1+t| + C \]

Substituting \(t = x^{1/6}\):

\[ = 3x^{1/3} - 6x^{1/6} + 6\log|1+x^{1/6}| + C \]

Solution:

Using partial fractions:

\[ \text{Let } \frac{5x}{(x+1)(x^2+9)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+9} \quad \dots(1) \]

\[ \Rightarrow 5x = A(x^2+9) + (Bx+C)(x+1) \] \[ \Rightarrow 5x = Ax^2 + 9A + Bx^2 + Bx + Cx + C \]

Equating coefficients of \(x^2\), \(x\), and the constant term:

\(A + B = 0\)
\(B + C = 5\)
\(9A + C = 0\)

Solving these equations, we obtain \(A = -\frac{1}{2}, B = \frac{1}{2}, \text{ and } C = \frac{9}{2}\).

From equation (1):

\[ \frac{5x}{(x+1)(x^2+9)} = \frac{-1}{2(x+1)} + \frac{\frac{x}{2} + \frac{9}{2}}{x^2+9} \]

Integrating both sides:

\[ \int \frac{5x}{(x+1)(x^2+9)} dx = \int \left\{ \frac{-1}{2(x+1)} + \frac{x+9}{2(x^2+9)} \right\} dx \] \[ = -\frac{1}{2}\log|x+1| + \frac{1}{2}\int \frac{x}{x^2+9}dx + \frac{9}{2}\int \frac{1}{x^2+9}dx \] \[ = -\frac{1}{2}\log|x+1| + \frac{1}{4}\log|x^2+9| + \frac{9}{2} \cdot \frac{1}{3} \tan^{-1}\frac{x}{3} \] \[ = -\frac{1}{2}\log|x+1| + \frac{1}{4}\log(x^2+9) + \frac{3}{2} \tan^{-1}\frac{x}{3} + C \]

Solution:

Let \(x - a = t \Rightarrow dx = dt\).

\[ \begin{aligned} \int \frac{\sin x}{\sin(x-a)} dx &= \int \frac{\sin(t+a)}{\sin t} dt \\ &= \int \frac{\sin t \cos a + \cos t \sin a}{\sin t} dt \\ &= \int (\cos a + \cot t \sin a) dt \\ &= t \cos a + \sin a \log|\sin t| + C_1 \\ &= (x-a)\cos a + \sin a \log|\sin(x-a)| + C_1 \\ &= x \cos a - a \cos a + \sin a \log|\sin(x-a)| + C_1 \end{aligned} \]

Combining the constants \(-a \cos a + C_1\) into a new constant \(C\):

\[ = \sin a \log|\sin(x-a)| + x \cos a + C \]

Solution:

Using the property \(e^{m \log x} = e^{\log x^m} = x^m\), we simplify the expression:

\[ \frac{e^{5 \log x} - e^{4 \log x}}{e^{3 \log x} - e^{2 \log x}} = \frac{x^5 - x^4}{x^3 - x^2} \]

Factorizing the numerator and denominator:

\[ = \frac{x^4(x-1)}{x^2(x-1)} = x^2 \]

Now integrate:

\[ \Rightarrow \int \frac{e^{5 \log x} - e^{4 \log x}}{e^{3 \log x} - e^{2 \log x}} dx = \int x^2 dx = \frac{x^3}{3} + C \]

Solution:

Let \(\sin x = t \Rightarrow \cos x dx = dt\).

Substituting into the integral:

\[ \Rightarrow \int \frac{\cos x}{\sqrt{4 - \sin^2 x}} dx = \int \frac{dt}{\sqrt{(2)^2 - (t)^2}} \]

Using the standard integral \(\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a}\):

\[ = \sin^{-1}\left(\frac{t}{2}\right) + C \] \[ = \sin^{-1}\left(\frac{\sin x}{2}\right) + C \]

Solution:

Simplifying the numerator using \(a^2 - b^2 = (a-b)(a+b)\):

\[ \sin^8 x - \cos^8 x = (\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x) \] \[ = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x)(\sin^4 x + \cos^4 x) \] \[ = -\cos 2x (1) (\sin^4 x + \cos^4 x) \]

Now simplifying the denominator:

\[ 1 - 2\sin^2 x \cos^2 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x \] \[ = \sin^4 x + \cos^4 x + 2\sin^2 x \cos^2 x - 2\sin^2 x \cos^2 x \] \[ = \sin^4 x + \cos^4 x \]

Substituting these back into the expression, the term \((\sin^4 x + \cos^4 x)\) cancels out:

\[ \frac{\sin^8 x - \cos^8 x}{1 - 2\sin^2 x \cos^2 x} = \frac{-\cos 2x (\sin^4 x + \cos^4 x)}{(\sin^4 x + \cos^4 x)} = -\cos 2x \]

Integrating the simplified function:

\[ \Rightarrow \int \frac{\sin^8 x - \cos^8 x}{1 - 2\sin^2 x \cos^2 x} dx = \int -\cos 2x dx = -\frac{\sin 2x}{2} + C \]

Solution:

Multiply and divide the numerator and denominator by \(\sin(a-b)\):

Express \((a-b)\) as \((x+a) - (x+b)\) in the numerator:

Splitting the terms:

Integrating \(\int \tan u \, du = -\log|\cos u|\):

Solution:

Rewrite the integral as:

Let \(x^4 = t\). Differentiating both sides:

Substitute these into the integral:

Substitute \(t = x^4\) back:

Solution:

Let \(e^x = t\). Then \(e^x dx = dt\).

The integral becomes:

Using partial fractions:

Integrating term by term:

Substituting \(t = e^x\) back:

Solution:

Using partial fraction decomposition:

Expanding and equating coefficients of \(x^3, x^2, x\) and constant terms:

• \(A+C = 0\)
• \(B+D = 0\)
• \(4A+C = 0\)
• \(4B+D = 1\)

Solving these equations gives:

The integral becomes:

Using the standard integral \(\int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1} \frac{x}{a}\):

Solution:

Simplify the term \(e^{\log \sin x} = \sin x\).

The integral becomes:

Let \(\cos x = t\). Then \(-\sin x \, dx = dt \Rightarrow \sin x \, dx = -dt\).

Substitute \(t = \cos x\) back:

Solution:

Simplify using logarithmic properties: \(e^{3 \log x} = e^{\log x^3} = x^3\).

The integral becomes:

Let \(x^4 + 1 = t\). Then \(4x^3 dx = dt \Rightarrow x^3 dx = \frac{1}{4} dt\).

Substitute \(t = x^4 + 1\) back:

Solution:

Let \(I = \int f'(ax+b)[f(ax+b)]^n dx\).

Let \(f(ax+b) = t\). Differentiating both sides with respect to \(x\):

Substitute these values into the integral:

Solution:

Expand \(\sin(x+\alpha)\) in the denominator:

Factor out \(\sin^4 x\) inside the square root:

Let \(\cos \alpha + \cot x \sin \alpha = t\). Differentiating:

Substituting into the integral:

Solution:

We know that \(\sin^{-1}\sqrt{x} + \cos^{-1}\sqrt{x} = \frac{\pi}{2}\). Thus, the denominator is \(\frac{\pi}{2}\).

Also, substitute \(\sin^{-1}\sqrt{x} = \frac{\pi}{2} - \cos^{-1}\sqrt{x}\) into the numerator:

Let \(I_1 = \int \cos^{-1}\sqrt{x} \, dx\).

Put \(\sqrt{x} = t \Rightarrow x = t^2 \Rightarrow dx = 2t \, dt\).

Using integration by parts:

Let \(1-t^2 = u \Rightarrow -2t dt = du \Rightarrow t dt = -du/2\).

Substitute back into equation (1) using \(t = \sqrt{x}\):

Simplifying further (as per textbook answer format):

Solution:

Let \( I = \int \frac{2 + \sin 2x}{1 + \cos 2x}e^x dx \)

Using the trigonometric identities \( \sin 2x = 2\sin x \cos x \) and \( 1 + \cos 2x = 2\cos^2 x \), we get:

\[ \begin{aligned} I &= \int \left( \frac{2 + 2\sin x \cos x}{2\cos^2 x} \right) e^x dx \\ &= \int \left( \frac{1 + \sin x \cos x}{\cos^2 x} \right) e^x dx \\ &= \int \left( \frac{1}{\cos^2 x} + \frac{\sin x \cos x}{\cos^2 x} \right) e^x dx \\ &= \int (\sec^2 x + \tan x)e^x dx \end{aligned} \]

Let \( f(x) = \tan x \Rightarrow f'(x) = \sec^2 x \).

We know that \( \int e^x [f(x) + f'(x)] dx = e^x f(x) + C \).

Therefore,

\[ I = e^x \tan x + C \]

Solution:

Let use partial fraction decomposition:

\[ \frac{x^2 + x + 1}{(x + 1)^2(x + 2)} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{C}{x + 2} \quad \dots(1) \]

Multiplying both sides by \( (x + 1)^2(x + 2) \):

\[ x^2 + x + 1 = A(x + 1)(x + 2) + B(x + 2) + C(x + 1)^2 \]

Expanding the terms:

\[ x^2 + x + 1 = A(x^2 + 3x + 2) + B(x + 2) + C(x^2 + 2x + 1) \]

Grouping coefficients:

\[ x^2 + x + 1 = (A + C)x^2 + (3A + B + 2C)x + (2A + 2B + C) \]

Equating the coefficients of \( x^2 \), \( x \), and the constant term, we obtain:

1. \( A + C = 1 \)
2. \( 3A + B + 2C = 1 \)
3. \( 2A + 2B + C = 1 \)

On solving these equations, we obtain \( A = -2 \), \( B = 1 \), and \( C = 3 \).

Substituting these values into equation (1):

\[ \frac{x^2 + x + 1}{(x + 1)^2(x + 2)} = \frac{-2}{x + 1} + \frac{1}{(x + 1)^2} + \frac{3}{x + 2} \]

Integrating both sides:

\[ \begin{aligned} \int \frac{x^2 + x + 1}{(x + 1)^2(x + 2)} dx &= -2 \int \frac{1}{x + 1} dx + \int (x + 1)^{-2} dx + 3 \int \frac{1}{x + 2} dx \\ &= -2 \log|x + 1| + \frac{(x + 1)^{-1}}{-1} + 3 \log|x + 2| + C \\ &= -2 \log|x + 1| - \frac{1}{x + 1} + 3 \log|x + 2| + C \end{aligned} \]

Solution:

Let \( x = \cos \theta \Rightarrow dx = -\sin \theta d\theta \).

Also, \( \theta = \cos^{-1} x \).

\[ \begin{aligned} I &= \int \tan^{-1} \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} (-\sin \theta d\theta) \\ &= -\int \tan^{-1} \sqrt{\frac{2\sin^2 \frac{\theta}{2}}{2\cos^2 \frac{\theta}{2}}} \sin \theta d\theta \\ &= -\int \tan^{-1} \left( \tan \frac{\theta}{2} \right) \sin \theta d\theta \\ &= -\int \frac{\theta}{2} \sin \theta d\theta \\ &= -\frac{1}{2} \int \theta \sin \theta d\theta \end{aligned} \]

Integrating by parts with \( u = \theta \) and \( dv = \sin \theta d\theta \):

\[ \begin{aligned} I &= -\frac{1}{2} \left[ \theta(-\cos \theta) - \int 1 \cdot (-\cos \theta) d\theta \right] \\ &= -\frac{1}{2} [-\theta \cos \theta + \sin \theta] \\ &= \frac{1}{2} \theta \cos \theta - \frac{1}{2} \sin \theta \end{aligned} \]

Substituting \( \theta = \cos^{-1} x \) and \( \sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - x^2} \):

\[ \begin{aligned} I &= \frac{1}{2} (\cos^{-1} x) \cdot x - \frac{1}{2} \sqrt{1 - x^2} + C \\ &= \frac{1}{2} \left( x \cos^{-1} x - \sqrt{1 - x^2} \right) + C \end{aligned} \]

Solution:

First, simplify the logarithmic term:

\[ \log(x^2 + 1) - 2\log x = \log(x^2 + 1) - \log x^2 = \log\left(\frac{x^2 + 1}{x^2}\right) = \log\left(1 + \frac{1}{x^2}\right) \]

Now rewrite the integral:

\[ \begin{aligned} I &= \int \frac{\sqrt{x^2 + 1}}{x^4} \log\left(1 + \frac{1}{x^2}\right) dx \\ &= \int \frac{\sqrt{x^2(1 + \frac{1}{x^2})}}{x^4} \log\left(1 + \frac{1}{x^2}\right) dx \\ &= \int \frac{x\sqrt{1 + \frac{1}{x^2}}}{x^4} \log\left(1 + \frac{1}{x^2}\right) dx \\ &= \int \frac{1}{x^3} \sqrt{1 + \frac{1}{x^2}} \log\left(1 + \frac{1}{x^2}\right) dx \end{aligned} \]

Let \( 1 + \frac{1}{x^2} = t \). Differentiating both sides:

\[ \frac{-2}{x^3} dx = dt \Rightarrow \frac{1}{x^3} dx = -\frac{1}{2} dt \]

Substituting into the integral:

\[ \begin{aligned} I &= \int \sqrt{t} \log t \left( -\frac{1}{2} \right) dt \\ &= -\frac{1}{2} \int t^{\frac{1}{2}} \log t dt \end{aligned} \]

Integrating by parts taking \( \log t \) as the first function:

\[ \begin{aligned} I &= -\frac{1}{2} \left[ \log t \cdot \frac{t^{3/2}}{3/2} - \int \frac{1}{t} \cdot \frac{t^{3/2}}{3/2} dt \right] \\ &= -\frac{1}{2} \left[ \frac{2}{3} t^{3/2} \log t - \frac{2}{3} \int t^{1/2} dt \right] \\ &= -\frac{1}{2} \left[ \frac{2}{3} t^{3/2} \log t - \frac{2}{3} \cdot \frac{t^{3/2}}{3/2} \right] \\ &= -\frac{1}{3} t^{3/2} \log t + \frac{1}{3} \cdot \frac{2}{3} t^{3/2} \\ &= -\frac{1}{3} t^{3/2} \log t + \frac{2}{9} t^{3/2} \\ &= -\frac{1}{3} t^{3/2} \left[ \log t - \frac{2}{3} \right] \end{aligned} \]

Substituting \( t = 1 + \frac{1}{x^2} \) back:

\[ I = -\frac{1}{3} \left(1 + \frac{1}{x^2}\right)^{\frac{3}{2}} \left[ \log\left(1 + \frac{1}{x^2}\right) - \frac{2}{3} \right] + C \]

Answer: \( e^{\frac{\pi}{2}} \)

Let \( I = \int_{\frac{\pi}{2}}^{\pi} e^x \left(\frac{1-\sin x}{1-\cos x}\right) dx \).

We know that \( 1 - \cos x = 2\sin^2\frac{x}{2} \) and \( \sin x = 2\sin\frac{x}{2}\cos\frac{x}{2} \).

Substituting these in the integrand:

\[ \begin{aligned} I &= \int_{\frac{\pi}{2}}^{\pi} e^x \left( \frac{1 - 2\sin\frac{x}{2}\cos\frac{x}{2}}{2\sin^2\frac{x}{2}} \right) dx \\ &= \int_{\frac{\pi}{2}}^{\pi} e^x \left( \frac{1}{2\sin^2\frac{x}{2}} - \frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\sin^2\frac{x}{2}} \right) dx \\ &= \int_{\frac{\pi}{2}}^{\pi} e^x \left( \frac{1}{2}\csc^2\frac{x}{2} - \cot\frac{x}{2} \right) dx \end{aligned} \]

Let \( f(x) = -\cot\frac{x}{2} \). Then \( f'(x) = -(-\csc^2\frac{x}{2} \cdot \frac{1}{2}) = \frac{1}{2}\csc^2\frac{x}{2} \).

The integral is of the form \( \int e^x [f(x) + f'(x)] dx = e^x f(x) \).

\[ \begin{aligned} I &= \left[ e^x \left(-\cot\frac{x}{2}\right) \right]_{\frac{\pi}{2}}^{\pi} \\ &= \left[ -e^{\pi} \cot\frac{\pi}{2} - \left( -e^{\frac{\pi}{2}} \cot\frac{\pi}{4} \right) \right] \\ &= \left[ -e^{\pi}(0) + e^{\frac{\pi}{2}}(1) \right] \\ &= e^{\frac{\pi}{2}} \end{aligned} \]

Answer: \( \frac{\pi}{8} \)

Let \( I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos^4 x + \sin^4 x} dx \).

Divide the numerator and denominator by \( \cos^4 x \):

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\frac{\sin x \cos x}{\cos^4 x}}{1 + \tan^4 x} dx = \int_{0}^{\frac{\pi}{4}} \frac{\tan x \sec^2 x}{1 + (\tan^2 x)^2} dx \]

Let \( t = \tan^2 x \). Then \( dt = 2 \tan x \sec^2 x \, dx \Rightarrow \frac{dt}{2} = \tan x \sec^2 x \, dx \).

Change limits:

• When \( x = 0 \), \( t = 0 \).
• When \( x = \frac{\pi}{4} \), \( t = 1 \).

\[ \begin{aligned} I &= \frac{1}{2} \int_{0}^{1} \frac{dt}{1 + t^2} \\ &= \frac{1}{2} \left[ \tan^{-1} t \right]_{0}^{1} \\ &= \frac{1}{2} (\tan^{-1} 1 - \tan^{-1} 0) \\ &= \frac{1}{2} \left( \frac{\pi}{4} \right) \\ &= \frac{\pi}{8} \end{aligned} \]

Answer: \( \frac{\pi}{6} \)

Let \( I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^2 x}{\cos^2 x + 4 \sin^2 x} dx \).

Using \( \cos^2 x = \frac{1+\cos 2x}{2} \) and \( \sin^2 x = \frac{1-\cos 2x}{2} \):

\[ \begin{aligned} I &= \int_{0}^{\frac{\pi}{2}} \frac{\frac{1+\cos 2x}{2}}{\frac{1+\cos 2x}{2} + 4\left(\frac{1-\cos 2x}{2}\right)} dx \\ &= \int_{0}^{\frac{\pi}{2}} \frac{1+\cos 2x}{1 + \cos 2x + 4 - 4\cos 2x} dx \\ &= \int_{0}^{\frac{\pi}{2}} \frac{1+\cos 2x}{5 - 3\cos 2x} dx \end{aligned} \]

Rewrite the numerator in terms of the denominator: \( 1+\cos 2x = -\frac{1}{3}(5-3\cos 2x) + \frac{8}{3} \).

\[ \begin{aligned} I &= \int_{0}^{\frac{\pi}{2}} \left( -\frac{1}{3} + \frac{8}{3(5 - 3\cos 2x)} \right) dx \\ &= \left[ -\frac{x}{3} \right]_{0}^{\frac{\pi}{2}} + \frac{8}{3} \int_{0}^{\frac{\pi}{2}} \frac{dx}{5 - 3\cos 2x} \\ &= -\frac{\pi}{6} + \frac{8}{3} I_1 \end{aligned} \]

Solving \( I_1 = \int_{0}^{\frac{\pi}{2}} \frac{dx}{5 - 3\cos 2x} \):

Put \( \tan x = t \), then \( dx = \frac{dt}{1+t^2} \) and \( \cos 2x = \frac{1-t^2}{1+t^2} \). Limits become \( 0 \to \infty \).

\[ \begin{aligned} I_1 &= \int_{0}^{\infty} \frac{1}{5 - 3\left(\frac{1-t^2}{1+t^2}\right)} \cdot \frac{dt}{1+t^2} \\ &= \int_{0}^{\infty} \frac{dt}{5(1+t^2) - 3(1-t^2)} \\ &= \int_{0}^{\infty} \frac{dt}{2 + 8t^2} = \frac{1}{2} \int_{0}^{\infty} \frac{dt}{1 + 4t^2} \\ &= \frac{1}{2} \cdot \frac{1}{2} \left[ \tan^{-1}(2t) \right]_{0}^{\infty} \\ &= \frac{1}{4} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{8} \end{aligned} \]

Substituting back: \( I = -\frac{\pi}{6} + \frac{8}{3}\left(\frac{\pi}{8}\right) = -\frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{6} \).

Answer: \( 2\sin^{-1}\left(\frac{\sqrt{3}-1}{2}\right) \)

Let \( I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x + \cos x}{\sqrt{\sin 2x}} dx \).

Put \( \sin x - \cos x = t \). Then \( (\cos x + \sin x)dx = dt \).

Squaring both sides: \( (\sin x - \cos x)^2 = t^2 \Rightarrow 1 - \sin 2x = t^2 \Rightarrow \sin 2x = 1 - t^2 \).

Change limits:

• \( x = \frac{\pi}{6} \Rightarrow t = \sin\frac{\pi}{6} - \cos\frac{\pi}{6} = \frac{1-\sqrt{3}}{2} \).
• \( x = \frac{\pi}{3} \Rightarrow t = \sin\frac{\pi}{3} - \cos\frac{\pi}{3} = \frac{\sqrt{3}-1}{2} \).

Let \( a = \frac{\sqrt{3}-1}{2} \). The limits are from \( -a \) to \( a \).

\[ \begin{aligned} I &= \int_{-a}^{a} \frac{dt}{\sqrt{1-t^2}} \\ &= \left[ \sin^{-1} t \right]_{-a}^{a} \\ &= \sin^{-1}(a) - \sin^{-1}(-a) \\ &= 2\sin^{-1}(a) \\ &= 2\sin^{-1}\left(\frac{\sqrt{3}-1}{2}\right) \end{aligned} \]

Answer: \( \frac{4\sqrt{2}}{3} \)

Rationalize the denominator by multiplying the numerator and denominator by \( \sqrt{1+x} + \sqrt{x} \):

\[ \begin{aligned} I &= \int_{0}^{1} \frac{\sqrt{1+x} + \sqrt{x}}{(\sqrt{1+x} - \sqrt{x})(\sqrt{1+x} + \sqrt{x})} dx \\ &= \int_{0}^{1} \frac{\sqrt{1+x} + \sqrt{x}}{(1+x) - x} dx \\ &= \int_{0}^{1} (\sqrt{1+x} + \sqrt{x}) dx \\ &= \int_{0}^{1} (1+x)^{1/2} dx + \int_{0}^{1} x^{1/2} dx \end{aligned} \]

Integrating:

\[ \begin{aligned} I &= \left[ \frac{2}{3}(1+x)^{3/2} \right]_{0}^{1} + \left[ \frac{2}{3}x^{3/2} \right]_{0}^{1} \\ &= \frac{2}{3} \left( 2^{3/2} - 1^{3/2} \right) + \frac{2}{3} \left( 1^{3/2} - 0 \right) \\ &= \frac{2}{3} (2\sqrt{2} - 1) + \frac{2}{3} \\ &= \frac{4\sqrt{2}}{3} - \frac{2}{3} + \frac{2}{3} \\ &= \frac{4\sqrt{2}}{3} \end{aligned} \]

Answer: \( \frac{1}{20} \log 3 \)

Let \( \sin x - \cos x = t \). Then \( (\cos x + \sin x)dx = dt \).

Also, \( \sin 2x = 1 - t^2 \). (Derived by squaring \( t = \sin x - \cos x \))

Change limits:

• \( x = 0 \Rightarrow t = -1 \).
• \( x = \frac{\pi}{4} \Rightarrow t = 0 \).

The denominator becomes \( 9 + 16(1-t^2) = 9 + 16 - 16t^2 = 25 - 16t^2 \).

\[ \begin{aligned} I &= \int_{-1}^{0} \frac{dt}{25 - 16t^2} \\ &= \frac{1}{16} \int_{-1}^{0} \frac{dt}{\frac{25}{16} - t^2} \\ &= \frac{1}{16} \int_{-1}^{0} \frac{dt}{\left(\frac{5}{4}\right)^2 - t^2} \end{aligned} \]

Using \( \int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| \):

\[ \begin{aligned} I &= \frac{1}{16} \cdot \frac{1}{2(\frac{5}{4})} \left[ \ln \left| \frac{\frac{5}{4} + t}{\frac{5}{4} - t} \right| \right]_{-1}^{0} \\ &= \frac{1}{16} \cdot \frac{2}{5} \left[ \ln \left| \frac{5+4t}{5-4t} \right| \right]_{-1}^{0} \\ &= \frac{1}{40} \left[ \ln(1) - \ln \left| \frac{1}{9} \right| \right] \\ &= \frac{1}{40} (0 - \ln(3^{-2})) \\ &= \frac{1}{40} (2 \ln 3) \\ &= \frac{1}{20} \ln 3 \end{aligned} \]

Answer: \( \frac{\pi}{2} - 1 \)

Let \( I = \int_{0}^{\frac{\pi}{2}} \sin 2x \tan^{-1}(\sin x) \, dx \).

Using \( \sin 2x = 2\sin x \cos x \):

\[ I = \int_{0}^{\frac{\pi}{2}} 2\sin x \cos x \tan^{-1}(\sin x) \, dx \]

Let \( \sin x = t \). Then \( \cos x \, dx = dt \).

Change limits:

• When \( x = 0 \), \( t = 0 \).
• When \( x = \frac{\pi}{2} \), \( t = 1 \).

Substituting these values:

\[ I = \int_{0}^{1} 2t \tan^{-1} t \, dt \]

Using integration by parts with \( u = \tan^{-1} t \) and \( dv = 2t \, dt \):

• \( du = \frac{1}{1+t^2} \, dt \)
• \( v = t^2 \)

\[ \begin{aligned} I &= \left[ t^2 \tan^{-1} t \right]_{0}^{1} - \int_{0}^{1} \frac{t^2}{1+t^2} \, dt \\ &= (1 \cdot \frac{\pi}{4} - 0) - \int_{0}^{1} \frac{1+t^2-1}{1+t^2} \, dt \\ &= \frac{\pi}{4} - \int_{0}^{1} \left( 1 - \frac{1}{1+t^2} \right) dt \\ &= \frac{\pi}{4} - \left[ t - \tan^{-1} t \right]_{0}^{1} \\ &= \frac{\pi}{4} - \left[ (1 - \frac{\pi}{4}) - (0 - 0) \right] \\ &= \frac{\pi}{4} - 1 + \frac{\pi}{4} \\ &= \frac{\pi}{2} - 1 \end{aligned} \]

Answer: \( \frac{19}{2} \)

Let \( I = \int_{1}^{4} (|x-1| + |x-2| + |x-3|) \, dx \).

We split the integral based on the critical points \( x=1, 2, 3 \):

• For \( 1 \le x < 2 \): \( |x-1| = x-1 \), \( |x-2| = -(x-2) \), \( |x-3| = -(x-3) \). Sum = \( x-1-x+2-x+3 = 4-x \).
• For \( 2 \le x < 3 \): \( |x-1| = x-1 \), \( |x-2| = x-2 \), \( |x-3| = -(x-3) \). Sum = \( x-1+x-2-x+3 = x \).
• For \( 3 \le x \le 4 \): All terms are positive. Sum = \( x-1+x-2+x-3 = 3x-6 \).

Thus,

\[ \begin{aligned} I &= \int_{1}^{2} (4-x) \, dx + \int_{2}^{3} x \, dx + \int_{3}^{4} (3x-6) \, dx \\ &= \left[ 4x - \frac{x^2}{2} \right]_{1}^{2} + \left[ \frac{x^2}{2} \right]_{2}^{3} + \left[ \frac{3x^2}{2} - 6x \right]_{3}^{4} \\ &= \left[ (8-2) - (4-\frac{1}{2}) \right] + \left[ \frac{9}{2} - 2 \right] + \left[ (24-24) - (\frac{27}{2} - 18) \right] \\ &= \left( 6 - \frac{7}{2} \right) + \frac{5}{2} + \left( 0 - (\frac{27-36}{2}) \right) \\ &= \frac{5}{2} + \frac{5}{2} + \frac{9}{2} \\ &= \frac{19}{2} \end{aligned} \]

Proof:

Let \( I = \int_{1}^{3} \frac{dx}{x^2(x+1)} \).

Using partial fractions: \( \frac{1}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} \).

\[ 1 = A x(x+1) + B(x+1) + C x^2 \]

• Put \( x = 0 \Rightarrow B = 1 \).
• Put \( x = -1 \Rightarrow C = 1 \).
• Comparing coefficients of \( x^2 \): \( A + C = 0 \Rightarrow A = -1 \).

Thus, the integrand is \( -\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x+1} \).

\[ \begin{aligned} I &= \int_{1}^{3} \left( -\frac{1}{x} + x^{-2} + \frac{1}{x+1} \right) dx \\ &= \left[ -\log|x| - \frac{1}{x} + \log|x+1| \right]_{1}^{3} \\ &= \left[ \log\left|\frac{x+1}{x}\right| - \frac{1}{x} \right]_{1}^{3} \\ &= \left( \log \frac{4}{3} - \frac{1}{3} \right) - \left( \log \frac{2}{1} - 1 \right) \\ &= \log \frac{4}{3} - \log 2 - \frac{1}{3} + 1 \\ &= \log \left( \frac{4}{3} \cdot \frac{1}{2} \right) + \frac{2}{3} \\ &= \frac{2}{3} + \log \frac{2}{3} \end{aligned} \]

Hence Proved.

Proof:

Let \( I = \int_{0}^{1} x e^x \, dx \).

Using integration by parts: \( u = x \), \( dv = e^x \, dx \).

• \( du = dx \)
• \( v = e^x \)

\[ \begin{aligned} I &= [x e^x]_{0}^{1} - \int_{0}^{1} e^x \, dx \\ &= (1 \cdot e^1 - 0) - [e^x]_{0}^{1} \\ &= e - (e^1 - e^0) \\ &= e - e + 1 \\ &= 1 \end{aligned} \]

Hence Proved.

Proof:

Let \( f(x) = x^{17} \cos^4 x \).

Check if the function is odd or even:

\[ f(-x) = (-x)^{17} \cos^4(-x) = -x^{17} \cos^4 x = -f(x) \]

Since \( f(x) \) is an odd function, the integral over the symmetric interval \( [-1, 1] \) is zero.

\[ \int_{-a}^{a} f(x) \, dx = 0 \text{ if } f(-x) = -f(x) \]

Therefore, \( \int_{-1}^{1} x^{17} \cos^4 x \, dx = 0 \).

Hence Proved.

Proof:

Let \( I = \int_{0}^{\frac{\pi}{2}} \sin^3 x \, dx = \int_{0}^{\frac{\pi}{2}} \sin^2 x \cdot \sin x \, dx \).

\[ I = \int_{0}^{\frac{\pi}{2}} (1 - \cos^2 x) \sin x \, dx \]

Let \( \cos x = t \Rightarrow -\sin x \, dx = dt \).

Change limits:

• When \( x = 0 \), \( t = 1 \).
• When \( x = \frac{\pi}{2} \), \( t = 0 \).

\[ \begin{aligned} I &= \int_{1}^{0} (1 - t^2) (-dt) \\ &= \int_{0}^{1} (1 - t^2) \, dt \\ &= \left[ t - \frac{t^3}{3} \right]_{0}^{1} \\ &= \left( 1 - \frac{1}{3} \right) - 0 \\ &= \frac{2}{3} \end{aligned} \]

Hence Proved.

Answer: \( \frac{\pi}{2} - 1 \)

Let \( I = \int_{0}^{\frac{\pi}{2}} \sin 2x \tan^{-1}(\sin x) \, dx \).

Using the identity \( \sin 2x = 2\sin x \cos x \):

\[ I = \int_{0}^{\frac{\pi}{2}} 2\sin x \cos x \tan^{-1}(\sin x) \, dx \]

Substitute \( \sin x = t \), then \( \cos x \, dx = dt \).

Limits change: When \( x=0, t=0 \). When \( x=\frac{\pi}{2}, t=1 \).

\[ I = \int_{0}^{1} 2t \tan^{-1} t \, dt \]

Using integration by parts, let \( u = \tan^{-1} t \) and \( dv = 2t \, dt \).

Then \( du = \frac{1}{1+t^2} dt \) and \( v = t^2 \).

\[ \begin{aligned} I &= \left[ t^2 \tan^{-1} t \right]_{0}^{1} - \int_{0}^{1} \frac{t^2}{1+t^2} dt \\ &= \left( 1^2 \cdot \frac{\pi}{4} - 0 \right) - \int_{0}^{1} \frac{1+t^2-1}{1+t^2} dt \\ &= \frac{\pi}{4} - \int_{0}^{1} \left( 1 - \frac{1}{1+t^2} \right) dt \\ &= \frac{\pi}{4} - \left[ t - \tan^{-1} t \right]_{0}^{1} \\ &= \frac{\pi}{4} - \left[ \left( 1 - \frac{\pi}{4} \right) - (0 - 0) \right] \\ &= \frac{\pi}{4} - 1 + \frac{\pi}{4} \\ &= \frac{\pi}{2} - 1 \end{aligned} \]

Answer: \( \frac{19}{2} \)

Let the integrand be \( f(x) = |x-1| + |x-2| + |x-3| \).

We split the integral at the critical points \( x = 2 \) and \( x = 3 \) within the interval \([1, 4]\).

\[ I = \int_{1}^{2} f(x) dx + \int_{2}^{3} f(x) dx + \int_{3}^{4} f(x) dx \]

• For \( 1 \le x < 2 \): \( f(x) = (x-1) - (x-2) - (x-3) = 4 - x \).
• For \( 2 \le x < 3 \): \( f(x) = (x-1) + (x-2) - (x-3) = x \).
• For \( 3 \le x \le 4 \): \( f(x) = (x-1) + (x-2) + (x-3) = 3x - 6 \).

Substituting these into the integral:

\[ \begin{aligned} I &= \int_{1}^{2} (4-x) dx + \int_{2}^{3} x dx + \int_{3}^{4} (3x-6) dx \\ &= \left[ 4x - \frac{x^2}{2} \right]_{1}^{2} + \left[ \frac{x^2}{2} \right]_{2}^{3} + \left[ \frac{3x^2}{2} - 6x \right]_{3}^{4} \\ &= \left[ (8 - 2) - (4 - 0.5) \right] + \left[ 4.5 - 2 \right] + \left[ (24 - 24) - (13.5 - 18) \right] \\ &= (6 - 3.5) + 2.5 + (0 - (-4.5)) \\ &= 2.5 + 2.5 + 4.5 \\ &= 9.5 = \frac{19}{2} \end{aligned} \]

Proof:

Using partial fractions, let \( \frac{1}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} \).

Solving for constants: \( 1 = A(x)(x+1) + B(x+1) + C(x^2) \).

• At \( x=0 \): \( 1 = B(1) \Rightarrow B=1 \).
• At \( x=-1 \): \( 1 = C(1) \Rightarrow C=1 \).
• Coeff of \( x^2 \): \( 0 = A + C \Rightarrow A = -1 \).

Thus, integrand is \( -\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x+1} \).

\[ \begin{aligned} I &= \int_{1}^{3} \left( -\frac{1}{x} + x^{-2} + \frac{1}{x+1} \right) dx \\ &= \left[ -\log|x| - \frac{1}{x} + \log|x+1| \right]_{1}^{3} \\ &= \left[ \log\left(\frac{x+1}{x}\right) - \frac{1}{x} \right]_{1}^{3} \\ &= \left( \log\frac{4}{3} - \frac{1}{3} \right) - \left( \log\frac{2}{1} - 1 \right) \\ &= \log\frac{4}{3} - \log 2 + \frac{2}{3} \\ &= \log\left( \frac{4}{3} \cdot \frac{1}{2} \right) + \frac{2}{3} \\ &= \frac{2}{3} + \log\frac{2}{3} \end{aligned} \]

Hence Proved.

Proof:

Using integration by parts: \( u = x, dv = e^x dx \Rightarrow du = dx, v = e^x \).

\[ \begin{aligned} I &= [xe^x]_{0}^{1} - \int_{0}^{1} e^x dx \\ &= (1\cdot e^1 - 0) - [e^x]_{0}^{1} \\ &= e - (e - 1) \\ &= 1 \end{aligned} \]

Hence Proved.

Proof:

Let \( f(x) = x^{17} \cos^4 x \).

Check symmetry: \( f(-x) = (-x)^{17} \cos^4(-x) = -x^{17} \cos^4 x = -f(x) \).

Since \( f(x) \) is an odd function and the limits \( -1 \) to \( 1 \) are symmetric around zero, the integral vanishes.

\[ \int_{-a}^{a} f(x) dx = 0 \quad (\text{if } f \text{ is odd}) \]

Therefore, the integral is 0.

Proof:

\[ \begin{aligned} I &= \int_{0}^{\frac{\pi}{2}} \sin^2 x \cdot \sin x \, dx \\ &= \int_{0}^{\frac{\pi}{2}} (1 - \cos^2 x) \sin x \, dx \end{aligned} \]

Let \( \cos x = t \Rightarrow -\sin x dx = dt \).

Limits: \( x=0 \to t=1 \), \( x=\frac{\pi}{2} \to t=0 \).

\[ \begin{aligned} I &= \int_{1}^{0} (1 - t^2) (-dt) \\ &= \int_{0}^{1} (1 - t^2) dt \\ &= \left[ t - \frac{t^3}{3} \right]_{0}^{1} \\ &= 1 - \frac{1}{3} \\ &= \frac{2}{3} \end{aligned} \]

Hence Proved.

Proof:

Let \( I = 2 \int_{0}^{\frac{\pi}{4}} \tan^3 x \, dx = 2 \int_{0}^{\frac{\pi}{4}} \tan x (\tan^2 x) \, dx \).

Using \( \tan^2 x = \sec^2 x - 1 \):

\[ I = 2 \int_{0}^{\frac{\pi}{4}} (\tan x \sec^2 x - \tan x) \, dx \]

Integrating term by term:

• \( \int \tan x \sec^2 x \, dx = \frac{\tan^2 x}{2} \) (put \( \tan x = t \))
• \( \int \tan x \, dx = \log|\sec x| \)

\[ \begin{aligned} I &= 2 \left[ \frac{\tan^2 x}{2} - \log|\sec x| \right]_{0}^{\frac{\pi}{4}} \\ &= \left[ \tan^2 x - 2\log|\sec x| \right]_{0}^{\frac{\pi}{4}} \\ &= (\tan^2 \frac{\pi}{4} - 2\log(\sec \frac{\pi}{4})) - (0 - 2\log 1) \\ &= 1^2 - 2\log(\sqrt{2}) - 0 \\ &= 1 - 2\left(\frac{1}{2}\log 2\right) \\ &= 1 - \log 2 \end{aligned} \]

Hence Proved.

Proof:

We use integration by parts taking \( u = \sin^{-1} x \) and \( dv = 1 \cdot dx \).

Then \( du = \frac{1}{\sqrt{1-x^2}} dx \) and \( v = x \).

\[ \begin{aligned} I &= [x \sin^{-1} x]_{0}^{1} - \int_{0}^{1} \frac{x}{\sqrt{1-x^2}} dx \\ &= (1 \cdot \frac{\pi}{2} - 0) - \int_{0}^{1} x(1-x^2)^{-1/2} dx \end{aligned} \]

For the integral part, let \( 1-x^2 = t \), then \( -2x dx = dt \). Limits: \( 1 \to 0 \).

\[ \begin{aligned} \int_{0}^{1} \frac{x}{\sqrt{1-x^2}} dx &= -\frac{1}{2} \int_{1}^{0} \frac{dt}{\sqrt{t}} \\ &= \frac{1}{2} \int_{0}^{1} t^{-1/2} dt \\ &= \frac{1}{2} [2\sqrt{t}]_{0}^{1} = 1 \end{aligned} \]

Substituting back:

\[ I = \frac{\pi}{2} - 1 \]

Hence Proved.

Answer: (A) \( \tan^{-1}(e^x) + C \)

Multiply numerator and denominator by \( e^x \):

\[ I = \int \frac{e^x dx}{e^{2x} + 1} \]

Let \( e^x = t \Rightarrow e^x dx = dt \).

\[ I = \int \frac{dt}{t^2 + 1} = \tan^{-1} t + C = \tan^{-1}(e^x) + C \]

Answer: (B) \( \log|\sin x + \cos x| + C \)

Using \( \cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x) \):

\[ \begin{aligned} I &= \int \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\sin x + \cos x)^2} dx \\ &= \int \frac{\cos x - \sin x}{\sin x + \cos x} dx \end{aligned} \]

Let \( \sin x + \cos x = t \). Then \( (\cos x - \sin x) dx = dt \).

\[ I = \int \frac{dt}{t} = \log|t| + C = \log|\sin x + \cos x| + C \]

Answer: (D) \( \frac{a+b}{2} \int_{a}^{b} f(x) dx \)

Let \( I = \int_{a}^{b} x f(x) dx \).

Using the property \( \int_{a}^{b} g(x) dx = \int_{a}^{b} g(a+b-x) dx \):

\[ I = \int_{a}^{b} (a+b-x) f(a+b-x) dx \]

Given \( f(a+b-x) = f(x) \), we substitute:

\[ \begin{aligned} I &= \int_{a}^{b} (a+b-x) f(x) dx \\ I &= (a+b) \int_{a}^{b} f(x) dx - \int_{a}^{b} x f(x) dx \\ I &= (a+b) \int_{a}^{b} f(x) dx - I \end{aligned} \]

Adding \( I \) to both sides:

\[ 2I = (a+b) \int_{a}^{b} f(x) dx \Rightarrow I = \frac{a+b}{2} \int_{a}^{b} f(x) dx \]