Class 12-NCERT Solutions-Chapter-07-Integrals-Ex 7.9

Solution:

Let \( x^2 + 1 = t \). Then \( 2x \, dx = dt \Rightarrow x \, dx = \frac{dt}{2} \).

Change limits:

• When \( x = 0, t = 0^2 + 1 = 1 \).
• When \( x = 1, t = 1^2 + 1 = 2 \).

\[ \begin{aligned} \int_{0}^{1} \frac{x}{x^2+1} \, dx &= \int_{1}^{2} \frac{1}{t} \cdot \frac{dt}{2} \\ &= \frac{1}{2} [\log t]_{1}^{2} \\ &= \frac{1}{2} (\log 2 - \log 1) \\ &= \frac{1}{2} (\log 2 - 0) \\ &= \frac{1}{2} \log 2 = \log \sqrt{2} \end{aligned} \]

Solution:

Rewrite integrand: \( \sqrt{\sin \phi} \cos^4 \phi \cos \phi = \sqrt{\sin \phi} (1-\sin^2 \phi)^2 \cos \phi \).

Let \( \sin \phi = t \). Then \( \cos \phi \, d\phi = dt \).

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

\[ \begin{aligned} I &= \int_{0}^{1} \sqrt{t} (1-t^2)^2 \, dt \\ &= \int_{0}^{1} t^{1/2} (1 - 2t^2 + t^4) \, dt \\ &= \int_{0}^{1} (t^{1/2} - 2t^{5/2} + t^{9/2}) \, dt \\ &= \left[ \frac{t^{3/2}}{3/2} - 2\frac{t^{7/2}}{7/2} + \frac{t^{11/2}}{11/2} \right]_{0}^{1} \\ &= \left( \frac{2}{3} - \frac{4}{7} + \frac{2}{11} \right) \\ &= 2 \left( \frac{1}{3} - \frac{2}{7} + \frac{1}{11} \right) \\ &= 2 \left( \frac{77 - 66 + 21}{231} \right) = 2 \left( \frac{32}{231} \right) \\ &= \frac{64}{231} \end{aligned} \]

Solution:

Let \( x = \tan \theta \). Then \( dx = \sec^2 \theta \, d\theta \).

Limits: \( x=0 \to \theta=0 \); \( x=1 \to \theta=\frac{\pi}{4} \).

Integrand: \( \sin^{-1}(\sin 2\theta) = 2\theta \).

\[ \begin{aligned} I &= \int_{0}^{\frac{\pi}{4}} 2\theta \sec^2 \theta \, d\theta \\ &= 2 \left[ \theta \tan \theta - \int \tan \theta \, d\theta \right]_{0}^{\frac{\pi}{4}} \quad (\text{Integration by Parts}) \\ &= 2 \left[ \theta \tan \theta - \log|\sec \theta| \right]_{0}^{\frac{\pi}{4}} \\ &= 2 \left[ \left( \frac{\pi}{4}(1) - \log \sqrt{2} \right) - (0 - 0) \right] \\ &= \frac{\pi}{2} - 2 \left( \frac{1}{2} \log 2 \right) \\ &= \frac{\pi}{2} - \log 2 \end{aligned} \]

Solution:

Let \( x+2 = t^2 \). Then \( dx = 2t \, dt \). Also \( x = t^2-2 \).

Limits: \( x=0 \to t=\sqrt{2} \); \( x=2 \to t=2 \).

\[ \begin{aligned} I &= \int_{\sqrt{2}}^{2} (t^2-2) \cdot t \cdot 2t \, dt \\ &= 2 \int_{\sqrt{2}}^{2} (t^4 - 2t^2) \, dt \\ &= 2 \left[ \frac{t^5}{5} - \frac{2t^3}{3} \right]_{\sqrt{2}}^{2} \\ &= 2 \left[ \left(\frac{32}{5} - \frac{16}{3}\right) - \left(\frac{4\sqrt{2}}{5} - \frac{4\sqrt{2}}{3}\right) \right] \\ &= 2 \left[ \frac{96-80}{15} - \sqrt{2} \left(\frac{12-20}{15}\right) \right] \\ &= 2 \left[ \frac{16}{15} + \frac{8\sqrt{2}}{15} \right] \\ &= \frac{16\sqrt{2}(\sqrt{2}+1)}{15} \end{aligned} \]

Solution:

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

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

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

Solution:

\( x+4-x^2 = -(x^2-x-4) = -[(x-1/2)^2 - 1/4 - 4] = -[(x-1/2)^2 - 17/4] = (\frac{\sqrt{17}}{2})^2 - (x-\frac{1}{2})^2 \).

Use formula \( \int \frac{dx}{a^2-x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| \).

\[ \begin{aligned} I &= \left[ \frac{1}{2(\frac{\sqrt{17}}{2})} \log \left| \frac{\frac{\sqrt{17}}{2} + (x-\frac{1}{2})}{\frac{\sqrt{17}}{2} - (x-\frac{1}{2})} \right| \right]_{0}^{2} \\ &= \frac{1}{\sqrt{17}} \left[ \log \left| \frac{\sqrt{17}+2x-1}{\sqrt{17}-2x+1} \right| \right]_{0}^{2} \\ &= \frac{1}{\sqrt{17}} \left[ \log \left| \frac{\sqrt{17}+3}{\sqrt{17}-3} \right| - \log \left| \frac{\sqrt{17}-1}{\sqrt{17}+1} \right| \right] \\ &= \frac{1}{\sqrt{17}} \log \left( \frac{(\sqrt{17}+3)(\sqrt{17}+1)}{(\sqrt{17}-3)(\sqrt{17}-1)} \right) \dots \text{(Simplify)} \end{aligned} \]

Solution:

\( x^2+2x+5 = (x+1)^2 + 4 = (x+1)^2 + 2^2 \).

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

Solution:

Let \( 2x = t \Rightarrow 2dx = dt \). Limits: \( 2 \to 4 \).

\[ \begin{aligned} I &= \int_{2}^{4} \left( \frac{1}{t/2} - \frac{1}{2(t/2)^2} \right) e^t \frac{dt}{2} \\ &= \frac{1}{2} \int_{2}^{4} \left( \frac{2}{t} - \frac{2}{t^2} \right) e^t \, dt \\ &= \int_{2}^{4} e^t \left( \frac{1}{t} - \frac{1}{t^2} \right) \, dt \end{aligned} \]

This is standard form \( \int e^t (f(t) + f'(t)) dt = e^t f(t) \). Here \( f(t) = 1/t \).

\[ I = \left[ \frac{e^t}{t} \right]_{2}^{4} = \frac{e^4}{4} - \frac{e^2}{2} = \frac{e^2(e^2-2)}{4} \]

Answer: (A)

Solution:

Take \( x^3 \) common from numerator: \( (x^3(x^{-2}-1))^{1/3} = x(x^{-2}-1)^{1/3} \).

Integrand becomes: \( \frac{x(x^{-2}-1)^{1/3}}{x^4} = \frac{(x^{-2}-1)^{1/3}}{x^3} = x^{-3}(x^{-2}-1)^{1/3} \).

Let \( x^{-2}-1 = t \Rightarrow -2x^{-3} dx = dt \Rightarrow x^{-3} dx = -dt/2 \).

Limits: \( x=1/3 \to t=9-1=8 \); \( x=1 \to t=1-1=0 \).

\[ \begin{aligned} I &= \int_{8}^{0} t^{1/3} (-dt/2) = \frac{1}{2} \int_{0}^{8} t^{1/3} dt \\ &= \frac{1}{2} \left[ \frac{3}{4} t^{4/3} \right]_{0}^{8} \\ &= \frac{3}{8} [ (2^3)^{4/3} - 0 ] \\ &= \frac{3}{8} (16) = 6 \end{aligned} \]

Answer: (B)

Solution:

By the First Fundamental Theorem of Calculus (Leibniz Rule):

\( \frac{d}{dx} \int_{0}^{x} g(t) \, dt = g(x) \).

Here \( g(t) = t \sin t \).

So, \( f'(x) = x \sin x \).