Class 12-NCERT Solutions-Chapter-07-Integrals-Ex 7.1

Solution:

The anti-derivative of \( \sin 2x \) is a function of \( x \) whose derivative is \( \sin 2x \).

We know that,

\[ \frac{d}{dx}(\cos 2x) = -2 \sin 2x \]

\[ \Rightarrow \sin 2x = -\frac{1}{2} \frac{d}{dx}(\cos 2x) \]

\[ \Rightarrow \sin 2x = \frac{d}{dx} \left( -\frac{1}{2} \cos 2x \right) \]

Therefore, the anti-derivative of \( \sin 2x \) is \( -\frac{1}{2} \cos 2x \).

Solution:

The anti-derivative of \( \cos 3x \) is a function of \( x \) whose derivative is \( \cos 3x \).

We know that,

\[ \frac{d}{dx}(\sin 3x) = 3 \cos 3x \]

\[ \Rightarrow \cos 3x = \frac{1}{3} \frac{d}{dx}(\sin 3x) \]

\[ \Rightarrow \cos 3x = \frac{d}{dx} \left( \frac{1}{3} \sin 3x \right) \]

Therefore, the anti-derivative of \( \cos 3x \) is \( \frac{1}{3} \sin 3x \).

Solution:

The anti-derivative of \( e^{2x} \) is the function of \( x \) whose derivative is \( e^{2x} \).

It is known that,

\[ \frac{d}{dx}(e^{2x}) = 2e^{2x} \]

\[ \Rightarrow e^{2x} = \frac{1}{2} \frac{d}{dx}(e^{2x}) \]

\[ \Rightarrow e^{2x} = \frac{d}{dx} \left( \frac{1}{2} e^{2x} \right) \]

Therefore, the anti-derivative of \( e^{2x} \) is \( \frac{1}{2} e^{2x} \).

Solution:

The anti-derivative of \( (ax + b)^2 \) is the function of \( x \) whose derivative is \( (ax + b)^2 \).

It is known that,

\[ \frac{d}{dx}(ax + b)^3 = 3a(ax + b)^2 \]

\[ \Rightarrow (ax + b)^2 = \frac{1}{3a} \frac{d}{dx}(ax + b)^3 \]

\[ \Rightarrow (ax + b)^2 = \frac{d}{dx} \left( \frac{1}{3a} (ax + b)^3 \right) \]

Therefore, the anti-derivative of \( (ax + b)^2 \) is \( \frac{1}{3a}(ax + b)^3 \).

Solution:

The anti-derivative of \( \sin 2x - 4e^{3x} \) is the function of \( x \) whose derivative is \( \sin 2x - 4e^{3x} \).

From previous questions, we know the anti-derivatives individually. Combining them:

The anti-derivative is:

\[ -\frac{1}{2} \cos 2x - \frac{4}{3} e^{3x} \]

Solution:

\[ \int (4e^{3x} + 1) \, dx \]

\[ = 4 \int e^{3x} \, dx + \int 1 \, dx \]

\[ = 4 \left( \frac{e^{3x}}{3} \right) + x + C \]

\[ = \frac{4}{3} e^{3x} + x + C \]

Solution:

\[ \int x^2 \left( 1 - \frac{1}{x^2} \right) \, dx \]

\[ = \int (x^2 - 1) \, dx \]

\[ = \int x^2 \, dx - \int 1 \, dx \]

\[ = \frac{x^3}{3} - x + C \]

Solution:

\[ \int (ax^2 + bx + c) \, dx \]

\[ = a \int x^2 \, dx + b \int x \, dx + c \int 1 \, dx \]

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

\[ = \frac{ax^3}{3} + \frac{bx^2}{2} + cx + C \]

Solution:

\[ \int (2x^2 + e^x) \, dx \]

\[ = 2 \int x^2 \, dx + \int e^x \, dx \]

\[ = \frac{2}{3} x^3 + e^x + C \]

Solution:

\[ \int \left( \sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 \, dx \]

Expanding the square:

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

\[ = \int x \, dx + \int \frac{1}{x} \, dx - 2 \int 1 \, dx \]

\[ = \frac{x^2}{2} + \log|x| - 2x + C \]

Solution:

\[ \int \frac{x^3 + 5x^2 - 4}{x^2} \, dx \]

Dividing each term by \( x^2 \):

\[ = \int \left( x + 5 - 4x^{-2} \right) \, dx \]

\[ = \int x \, dx + 5 \int 1 \, dx - 4 \int x^{-2} \, dx \]

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

\[ = \frac{x^2}{2} + 5x + \frac{4}{x} + C \]

Solution:

\[ \int \frac{x^3 + 3x + 4}{\sqrt{x}} \, dx \]

\[ = \int \left( x^{3 - \frac{1}{2}} + 3x^{1 - \frac{1}{2}} + 4x^{-\frac{1}{2}} \right) \, dx \]

\[ = \int \left( x^{\frac{5}{2}} + 3x^{\frac{1}{2}} + 4x^{-\frac{1}{2}} \right) \, dx \]

\[ = \frac{x^{\frac{7}{2}}}{\frac{7}{2}} + 3 \left( \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right) + 4 \left( \frac{x^{\frac{1}{2}}}{\frac{1}{2}} \right) + C \]

\[ = \frac{2}{7} x^{\frac{7}{2}} + 2x^{\frac{3}{2}} + 8\sqrt{x} + C \]

Solution:

Factorizing the numerator:

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

Therefore,

\[ \int \frac{(x^2 + 1)(x - 1)}{x - 1} \, dx \]

\[ = \int (x^2 + 1) \, dx \]

\[ = \frac{x^3}{3} + x + C \]

Solution:

\[ \int (1 - x)\sqrt{x} \, dx \]

Multiply \( \sqrt{x} \) (which is \( x^{1/2} \)) inside the bracket:

\[ = \int (x^{1/2} - x \cdot x^{1/2}) \, dx \]

\[ = \int (x^{1/2} - x^{3/2}) \, dx \]

Now, integrate each term:

\[ = \frac{x^{3/2}}{3/2} - \frac{x^{5/2}}{5/2} + C \]

\[ = \frac{2}{3}x^{3/2} - \frac{2}{5}x^{5/2} + C \]

Solution:

\[ \int \sqrt{x}(3x^2 + 2x + 3) \, dx \]

Expand the expression by multiplying \( \sqrt{x} \) (or \( x^{1/2} \)):

\[ = \int (3x^{2 + \frac{1}{2}} + 2x^{1 + \frac{1}{2}} + 3x^{\frac{1}{2}}) \, dx \]

\[ = \int (3x^{5/2} + 2x^{3/2} + 3x^{1/2}) \, dx \]

Integrate term by term:

\[ = 3\left( \frac{x^{7/2}}{7/2} \right) + 2\left( \frac{x^{5/2}}{5/2} \right) + 3\left( \frac{x^{3/2}}{3/2} \right) + C \]

\[ = 3\left( \frac{2}{7}x^{7/2} \right) + 2\left( \frac{2}{5}x^{5/2} \right) + 3\left( \frac{2}{3}x^{3/2} \right) + C \]

\[ = \frac{6}{7}x^{7/2} + \frac{4}{5}x^{5/2} + 2x^{3/2} + C \]

Solution:

\[ \int (2x - 3\cos x + e^x) \, dx \]

\[ = 2 \int x \, dx - 3 \int \cos x \, dx + \int e^x \, dx \]

\[ = 2\left( \frac{x^2}{2} \right) - 3(\sin x) + e^x + C \]

\[ = x^2 - 3\sin x + e^x + C \]

Solution:

\[ \int (2x^2 - 3\sin x + 5x^{1/2}) \, dx \]

\[ = 2 \int x^2 \, dx - 3 \int \sin x \, dx + 5 \int x^{1/2} \, dx \]

\[ = 2\left( \frac{x^3}{3} \right) - 3(-\cos x) + 5\left( \frac{x^{3/2}}{3/2} \right) + C \]

\[ = \frac{2}{3}x^3 + 3\cos x + 5\left( \frac{2}{3} \right)x^{3/2} + C \]

\[ = \frac{2}{3}x^3 + 3\cos x + \frac{10}{3}x^{3/2} + C \]

Solution:

\[ \int \sec x(\sec x + \tan x) \, dx \]

\[ = \int (\sec^2 x + \sec x \tan x) \, dx \]

We know that \( \int \sec^2 x \, dx = \tan x \) and \( \int \sec x \tan x \, dx = \sec x \).

\[ = \tan x + \sec x + C \]

Solution:

\[ \int \frac{\sec^2 x}{\text{cosec}^2 x} \, dx \]

Convert to sin and cos:

\[ = \int \frac{1/\cos^2 x}{1/\sin^2 x} \, dx = \int \frac{\sin^2 x}{\cos^2 x} \, dx \]

\[ = \int \tan^2 x \, dx \]

Use the identity \( \tan^2 x = \sec^2 x - 1 \):

\[ = \int (\sec^2 x - 1) \, dx \]

\[ = \int \sec^2 x \, dx - \int 1 \, dx \]

\[ = \tan x - x + C \]

Solution:

\[ \int \frac{2 - 3\sin x}{\cos^2 x} \, dx \]

Split the fraction:

\[ = \int \left( \frac{2}{\cos^2 x} - \frac{3\sin x}{\cos^2 x} \right) \, dx \]

\[ = \int \left( 2\sec^2 x - 3\tan x \sec x \right) \, dx \]

\[ = 2 \int \sec^2 x \, dx - 3 \int \sec x \tan x \, dx \]

\[ = 2\tan x - 3\sec x + C \]

Options:

• (A) \( \frac{1}{3}x^{1/3} + 2x^{1/2} + C \)
• (B) \( \frac{2}{3}x^{2/3} + \frac{1}{2}x^2 + C \)
• (C) \( \frac{2}{3}x^{3/2} + 2x^{1/2} + C \)
• (D) \( \frac{3}{2}x^{3/2} + \frac{1}{2}x^{1/2} + C \)

Solution:

\[ \int \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right) \, dx \]

\[ = \int \left( x^{1/2} + x^{-1/2} \right) \, dx \]

\[ = \frac{x^{3/2}}{3/2} + \frac{x^{1/2}}{1/2} + C \]

\[ = \frac{2}{3}x^{3/2} + 2x^{1/2} + C \]

Correct Option: (C)

Options:

• (A) \( x^4 + \frac{1}{x^3} - \frac{129}{8} \)
• (B) \( x^3 + \frac{1}{x^4} + \frac{129}{8} \)
• (C) \( x^4 + \frac{1}{x^3} + \frac{129}{8} \)
• (D) \( x^3 + \frac{1}{x^4} - \frac{129}{8} \)

Solution:

It is given that \( \frac{d}{dx} f(x) = 4x^3 - \frac{3}{x^4} \)

Therefore, the anti-derivative of \( 4x^3 - \frac{3}{x^4} = f(x) \)

\[ \Rightarrow f(x) = \int \left( 4x^3 - \frac{3}{x^4} \right) \, dx \]

\[ f(x) = 4 \int x^3 \, dx - 3 \int x^{-4} \, dx \]

\[ f(x) = 4 \left( \frac{x^4}{4} \right) - 3 \left( \frac{x^{-3}}{-3} \right) + C \]

\[ f(x) = x^4 + x^{-3} + C \]

\[ \Rightarrow f(x) = x^4 + \frac{1}{x^3} + C \]

Given that \( f(2) = 0 \):

\[ f(2) = (2)^4 + \frac{1}{(2)^3} + C = 0 \]

\[ \Rightarrow 16 + \frac{1}{8} + C = 0 \]

\[ \Rightarrow C = -\left( 16 + \frac{1}{8} \right) \]

\[ \Rightarrow C = -\frac{129}{8} \]

Substituting \( C \) back into the function:

\[ f(x) = x^4 + \frac{1}{x^3} - \frac{129}{8} \]

Correct Option: (A)