NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10 Integrals

Exercise 7.10: Definite Integrals by Substitution

Detailed step-by-step solutions for NCERT Class 12 Mathematics, Chapter 7, Exercise 7.10. These problems focus on evaluating definite integrals using the method of substitution.

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Question 1. Evaluate the integral: \( \int_{0}^{1} \frac{x}{x^2 + 1} \, dx \)

Step 1: Substitution Let \( x^2 + 1 = t \).
Differentiating both sides: \( 2x \, dx = dt \implies x \, dx = \frac{dt}{2} \).

Step 2: Change Limits When \( x = 0 \), \( t = 0^2 + 1 = 1 \).
When \( x = 1 \), \( t = 1^2 + 1 = 2 \).

Step 3: Integrate Substituting values into the integral: \[ \begin{align} I &= \int_{1}^{2} \frac{1}{t} \cdot \frac{dt}{2} \\ &= \frac{1}{2} \int_{1}^{2} \frac{dt}{t} \\ &= \frac{1}{2} [\log |t|]_{1}^{2} \\ &= \frac{1}{2} (\log 2 - \log 1) \end{align} \] Since \( \log 1 = 0 \): \[ I = \frac{1}{2} \log 2 \]

Question 2. Evaluate the integral: \( \int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos^5 \phi \, d\phi \)

Step 1: Simplify the Integrand Rewrite \( \cos^5 \phi \) as \( \cos \phi \cdot (\cos^2 \phi)^2 \): \[ I = \int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} (1 - \sin^2 \phi)^2 \cos \phi \, d\phi \]

Step 2: Substitution Let \( \sin \phi = t \).
Then \( \cos \phi \, d\phi = dt \).
Limits: When \( \phi = 0, t = 0 \). When \( \phi = \frac{\pi}{2}, t = 1 \).

Step 3: Integrate \[ \begin{align} I &= \int_{0}^{1} \sqrt{t} (1 - t^2)^2 \, dt \\ &= \int_{0}^{1} t^{1/2} (1 - 2t^2 + t^4) \, dt \\ &= \int_{0}^{1} (t^{1/2} - 2t^{5/2} + t^{9/2}) \, dt \\ &= \left[ \frac{2}{3}t^{3/2} - 2\left(\frac{2}{7}\right)t^{7/2} + \frac{2}{11}t^{11/2} \right]_{0}^{1} \\ &= \left( \frac{2}{3} - \frac{4}{7} + \frac{2}{11} \right) - 0 \\ &= \frac{154 - 132 + 42}{231} \\ &= \frac{64}{231} \end{align} \]

Question 3. Evaluate the integral: \( \int_{0}^{1} \sin^{-1} \left( \frac{2x}{1+x^2} \right) \, dx \)

Step 1: Substitution Let \( x = \tan \theta \implies dx = \sec^2 \theta \, d\theta \).
Limits: When \( x=0, \theta=0 \). When \( x=1, \theta=\frac{\pi}{4} \).

Step 2: Simplify Integrand \( \frac{2x}{1+x^2} = \frac{2\tan\theta}{1+\tan^2\theta} = \sin 2\theta \).
So, \( \sin^{-1}(\sin 2\theta) = 2\theta \).

Step 3: Integration by Parts \[ I = \int_{0}^{\frac{\pi}{4}} 2\theta \sec^2 \theta \, d\theta \] Using integration by parts: \( \int u v' = uv - \int u'v \).
Take \( u = \theta \) and \( v' = \sec^2 \theta \). Then \( u' = 1 \) and \( v = \tan \theta \). \[ \begin{align} I &= 2 \left( [\theta \tan \theta]_{0}^{\frac{\pi}{4}} - \int_{0}^{\frac{\pi}{4}} \tan \theta \, d\theta \right) \\ &= 2 \left( \left(\frac{\pi}{4} \cdot 1 - 0\right) - [\log |\sec \theta|]_{0}^{\frac{\pi}{4}} \right) \\ &= 2 \left( \frac{\pi}{4} - (\log \sqrt{2} - \log 1) \right) \\ &= \frac{\pi}{2} - 2 \log (2^{1/2}) \\ &= \frac{\pi}{2} - \log 2 \end{align} \]

Question 4. Evaluate the integral: \( \int_{0}^{2} x\sqrt{x+2} \, dx \)

Step 1: Substitution Let \( x+2 = t^2 \implies dx = 2t \, dt \). Also \( x = t^2 - 2 \).
Limits: When \( x=0, t=\sqrt{2} \). When \( x=2, t=2 \).

Step 2: Integrate \[ \begin{align} I &= \int_{\sqrt{2}}^{2} (t^2 - 2) \cdot t \cdot 2t \, dt \\ &= 2 \int_{\sqrt{2}}^{2} (t^4 - 2t^2) \, dt \\ &= 2 \left[ \frac{t^5}{5} - \frac{2t^3}{3} \right]_{\sqrt{2}}^{2} \end{align} \]

Step 3: Evaluate Limits Upper limit (\(t=2\)): \( 2(\frac{32}{5} - \frac{16}{3}) = 2(\frac{96-80}{15}) = \frac{32}{15} \).
Lower limit (\(t=\sqrt{2}\)): \( 2(\frac{4\sqrt{2}}{5} - \frac{4\sqrt{2}}{3}) = 8\sqrt{2}(\frac{1}{5}-\frac{1}{3}) = 8\sqrt{2}(\frac{-2}{15}) = -\frac{16\sqrt{2}}{15} \).
Result: \[ I = \frac{32}{15} - \left( -\frac{16\sqrt{2}}{15} \right) = \frac{16\sqrt{2}(\sqrt{2}+1)}{15} \]

Question 5. Evaluate the integral: \( \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos^2 x} \, dx \)

Step 1: Substitution Let \( \cos x = t \implies -\sin x \, dx = dt \implies \sin x \, dx = -dt \).
Limits: \( x=0 \to t=1 \); \( x=\frac{\pi}{2} \to t=0 \).

Step 2: Integrate \[ \begin{align} I &= \int_{1}^{0} \frac{-dt}{1+t^2} \\ &= \int_{0}^{1} \frac{dt}{1+t^2} \\ &= [\tan^{-1} t]_{0}^{1} \\ &= \tan^{-1}(1) - \tan^{-1}(0) \\ &= \frac{\pi}{4} \end{align} \]

Question 6. Evaluate the integral: \( \int_{0}^{2} \frac{dx}{x+4-x^2} \)

Step 1: Complete the Square Denominator: \( -(x^2 - x - 4) = -[(x - \frac{1}{2})^2 - \frac{1}{4} - 4] = -[(x-\frac{1}{2})^2 - \frac{17}{4}] \)
\( = \left( \frac{\sqrt{17}}{2} \right)^2 - \left( x - \frac{1}{2} \right)^2 \).

Step 2: Apply Formula Use \( \int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| \).
Here \( a = \frac{\sqrt{17}}{2} \). \[ \begin{align} I &= \frac{1}{2(\frac{\sqrt{17}}{2})} \left[ \log \left| \frac{\frac{\sqrt{17}}{2} + (x-\frac{1}{2})}{\frac{\sqrt{17}}{2} - (x-\frac{1}{2})} \right| \right]_{0}^{2} \\ &= \frac{1}{\sqrt{17}} \left[ \log \left| \frac{\sqrt{17} + 2x - 1}{\sqrt{17} - 2x + 1} \right| \right]_{0}^{2} \end{align} \]

Step 3: Evaluate Limits Upper limit (\(x=2\)): \( \log \left| \frac{\sqrt{17} + 3}{\sqrt{17} - 3} \right| \).
Lower limit (\(x=0\)): \( \log \left| \frac{\sqrt{17} - 1}{\sqrt{17} + 1} \right| \).
Combining logarithms leads to: \[ I = \frac{1}{\sqrt{17}} \log \left( \frac{21 + 5\sqrt{17}}{4} \right) \]

Question 7. Evaluate the integral: \( \int_{-1}^{1} \frac{dx}{x^2+2x+5} \)

Step 1: Complete the Square \( x^2 + 2x + 5 = (x^2 + 2x + 1) + 4 = (x+1)^2 + 2^2 \).

Step 2: Integrate Using \( \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) \): \[ \begin{align} I &= \int_{-1}^{1} \frac{dx}{(x+1)^2 + 2^2} \\ &= \left[ \frac{1}{2} \tan^{-1}\left(\frac{x+1}{2}\right) \right]_{-1}^{1} \end{align} \]

Step 3: Evaluate \[ \begin{align} I &= \frac{1}{2} \left( \tan^{-1}(1) - \tan^{-1}(0) \right) \\ &= \frac{1}{2} \left( \frac{\pi}{4} - 0 \right) \\ &= \frac{\pi}{8} \end{align} \]

Question 8. Evaluate the integral: \( \int_{1}^{2} \left( \frac{1}{x} - \frac{1}{2x^2} \right) e^{2x} \, dx \)

Step 1: Substitution Let \( 2x = t \implies x = t/2 \implies dx = dt/2 \).
Limits: \( x=1 \to t=2 \); \( x=2 \to t=4 \).

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

Step 3: Apply Property We know \( \int e^x [f(x) + f'(x)] dx = e^x f(x) \).
Here \( f(t) = 1/t \) and \( f'(t) = -1/t^2 \). \[ \begin{align} I &= \left[ \frac{e^t}{t} \right]_{2}^{4} \\ &= \frac{e^4}{4} - \frac{e^2}{2} \\ &= \frac{e^2(e^2 - 2)}{4} \end{align} \]

Question 9. The value of the integral \( \int_{\frac{1}{3}}^{1} \frac{(x-x^3)^{\frac{1}{3}}}{x^4} \, dx \) is:

Step 1: Simplify Numerator Factor \( x^3 \) out of the bracket: \[ (x-x^3)^{1/3} = [x^3(x^{-2}-1)]^{1/3} = x(x^{-2}-1)^{1/3} \]

Step 2: Rewrite Integral \[ I = \int_{\frac{1}{3}}^{1} \frac{x(x^{-2}-1)^{1/3}}{x^4} \, dx = \int_{\frac{1}{3}}^{1} (x^{-2}-1)^{1/3} \cdot x^{-3} \, dx \]

Step 3: Substitution Let \( x^{-2} - 1 = t \).
Diff: \( -2x^{-3} \, dx = dt \implies x^{-3} \, dx = -\frac{1}{2} dt \).
Limits: \( x=\frac{1}{3} \implies t = 9-1=8 \); \( x=1 \implies t = 1-1=0 \).

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

Correct Option: A

Question 10. If \( f(x) = \int_{0}^{x} t \sin t \, dt \), then \( f'(x) \) is:

Step 1: Applying Fundamental Theorem of Calculus (Leibniz Rule) The derivative of an integral function \( F(x) = \int_{a}^{x} g(t) \, dt \) is simply \( g(x) \). \[ \frac{d}{dx} \left( \int_{0}^{x} t \sin t \, dt \right) = x \sin x \]

Alternatively, integrating by parts first: \[ f(x) = [-t \cos t + \sin t]_0^x = -x \cos x + \sin x \] Differentiating this: \[ f'(x) = -(\cos x - x \sin x) + \cos x = x \sin x \]

Correct Option: B