CBSE Class 12 Mathematics Indefinite And Definite Integrals Worksheet Set 08

CBSE Class 12 Mathematics Indefinite and Definite Integrals Worksheet (8). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects. 

Question. Evaluate:
\[ I = \int_{1}^{3} \frac{1}{x^2(x+1)} \, dx \]
Answer:
Type: partial fraction
let \( \frac{1}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} \)
\[ 1 = A(x)(x+1) + B(x+1) + C(x^2) \]
\[ 1 = A(x^2 + x) + B(x+1) + C(x^2) \]
Comp. coeff. of \( x^2 \), \( x \) & constant:
\( 0 = A + C \)
\( 0 = A + B \)
\( 1 = B \)

Solving these equations we get:
\( B = 1, A = -1, C = 1 \)

\[ \therefore I = \int_{1}^{3} \left( -\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x+1} \right) \, dx \]
\[ = \left[ -\log x - \frac{1}{x} + \log(x+1) \right]_{1}^{3} \]
\[ = \left[ \left( -\log 3 - \frac{1}{3} + \log 4 \right) - \left( -\log 1 - 1 + \log 2 \right) \right] \]
\[ = -\log 3 - \frac{1}{3} + \log 4 + 0 + 1 - \log 2 \]
\[ = -\log 3 + \frac{2}{3} + 2\log 2 - \log 2 \]
\[ = \frac{2}{3} + \log 2 - \log 3 \]
\[ I = \frac{2}{3} + \log\left(\frac{2}{3}\right) \quad \text{Ans.} \]

Question. Evaluate:
\[ I = \int_{0}^{\frac{\pi}{4}} \sqrt{1 - \sin 2x} \, dx \]
Answer:
\[ I = \int_{0}^{\frac{\pi}{4}} \sqrt{\sin^2 x + \cos^2 x - 2\sin x \cos x} \, dx \]
\[ I = \int_{0}^{\frac{\pi}{4}} \sqrt{(\cos x - \sin x)^2} \, dx \quad \dots \left\{ \begin{aligned} 0 < x < \frac{\pi}{4} \\ \cos x > \sin x \end{aligned} \right\} \]
\[ I = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx \]
\[ I = [ \sin x + \cos x ]_{0}^{\frac{\pi}{4}} \]
\[ I = \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) - (0 + 1) \]
\[ I = \sqrt{2} - 1 \quad \text{ans.} \]

Question. Evaluate:
\[ I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{1 - \sin(2x)} \, dx \]
Answer:
\[ I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{1 - \sin(2x)} \, dx \]
\[ I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{\sin^2 x + \cos^2 x - 2\sin x \cos x} \, dx \]
\[ I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{(\sin x - \cos x)^2} \, dx \quad \dots \left[ \begin{aligned} \frac{\pi}{4} < x < \frac{\pi}{2} \\ \sin x > \cos x \end{aligned} \right] \]
\[ I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin x - \cos x) \, dx \]
\[ I = [ -\cos x - \sin x ]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \]
\[ I = -[ \cos x + \sin x ]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \]
\[ I = -\left[ (0 + 1) - \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) \right] \]
\[ I = -1 + \sqrt{2} \]
\[ I = \sqrt{2} - 1 \quad \text{Ans.} \]

Question. Evaluate:
\[ I = \int_{0}^{2\pi} e^x \cdot \sin\left(\frac{\pi}{4} + \frac{x}{2}\right) \, dx \]
Answer:
{type: Repeats after integrating by parts twice}
Applying integration by parts:
\[ I = \left[ \sin\left(\frac{\pi}{4} + \frac{x}{2}\right) \cdot e^x \right]_{0}^{2\pi} - \int_{0}^{2\pi} \frac{1}{2} \cos\left(\frac{\pi}{4} + \frac{x}{2}\right) \cdot e^x \, dx \]
\[ I = \left[ \sin\left(\pi + \frac{\pi}{4}\right) \cdot e^{2\pi} - \sin\left(\frac{\pi}{4}\right) \cdot e^0 \right] - \frac{1}{2} \int_{0}^{2\pi} e^x \cdot \cos\left(\frac{\pi}{4} + \frac{x}{2}\right) \, dx \]
\[ I = \left[ -\frac{1}{\sqrt{2}} \cdot e^{2\pi} - \frac{1}{\sqrt{2}} \right] - \frac{1}{2} \left[ \left[ \cos\left(\frac{x}{2} + \frac{\pi}{4}\right) \cdot e^x \right]_{0}^{2\pi} + \frac{1}{2} \int_{0}^{2\pi} e^x \cdot \sin\left(\frac{x}{2} + \frac{\pi}{4}\right) \, dx \right] \]
\[ I = \left( \frac{-e^{2\pi} - 1}{\sqrt{2}} \right) - \frac{1}{2} \left[ \cos\left(\pi + \frac{\pi}{4}\right) \cdot e^{2\pi} - \cos\left(\frac{\pi}{4}\right) e^0 \right] - \frac{1}{4} \int_{0}^{2\pi} e^x \cdot \sin\left(\frac{x}{2} + \frac{\pi}{4}\right) \, dx \]
\[ I = \frac{-e^{2\pi} - 1}{\sqrt{2}} - \frac{1}{2} \left[ -\frac{1}{\sqrt{2}} \cdot e^{2\pi} - \frac{1}{\sqrt{2}} \right] - \frac{1}{4} I \]
\[ I + \frac{1}{4} I = \frac{-e^{2\pi} - 1}{\sqrt{2}} + \frac{1}{2\sqrt{2}} (e^{2\pi} + 1) \]
\[ \frac{5I}{4} = \frac{-2e^{2\pi} - 2 + e^{2\pi} + 1}{2\sqrt{2}} \]
\[ \frac{5I}{4} = \frac{-(e^{2\pi} + 1)}{2\sqrt{2}} \]
\[ I = -\frac{2}{5\sqrt{2}} (e^{2\pi} + 1) \]
\[ I = -\frac{\sqrt{2}}{5} (e^{2\pi} + 1) \quad \text{ans.} \]

Question. Evaluate:
\[ I = \int_{e}^{e^2} \left( \frac{1}{\log x} - \frac{1}{(\log x)^2} \right) \, dx \]
Answer:
Put \( \log x = t \text{ when } x = e \Rightarrow t = 1 \)
\( x = e^t \text{ when } x = e^2 \Rightarrow t = 2 \)
\( dx = e^t \, dt \)

\[ I = \int_{1}^{2} e^t \left( \frac{1}{t} - \frac{1}{t^2} \right) \, dt \]
\[ = \int_{1}^{2} e^t \cdot \frac{1}{t} \, dt - \int_{1}^{2} e^t \cdot \frac{1}{t^2} \, dt \]
\[ = \left[ e^t \cdot \frac{1}{t} \right]_{1}^{2} + \int_{1}^{2} \frac{1}{t^2} \cdot e^t \, dt - \int_{1}^{2} e^t \cdot \frac{1}{t^2} \, dt \]
\[ = \left[ e^2 \cdot \frac{1}{2} \right] - [e] \]
\[ I = \frac{e^2}{2} - e \quad \text{ans.} \]

Question. Evaluate:
\[ I = \int_{0}^{\pi} \left( \sin^2 \frac{x}{2} - \cos^2 \frac{x}{2} \right) \, dx \]
Answer:
\[ I = \int_{0}^{\pi} \left( \sin^2 \frac{x}{2} - \cos^2 \frac{x}{2} \right) \, dx \]
\[ = \int_{0}^{\pi} -\left( \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} \right) \, dx \]
\[ = \int_{0}^{\pi} -\cos x \, dx \quad \dots [\cos 2\theta = \cos^2\theta - \sin^2\theta] \]
\[ = -[ \sin x ]_{0}^{\pi} \]
\[ = -(\sin\pi - \sin 0) \]
\[ = 0 \quad \text{Ans.} \]

Question. Evaluate:
\[ I = \int_{0}^{\infty} \frac{x}{(1+x)(1+x^2)} \, dx \]
Answer:
\[ \frac{\pi}{4} \]
Hint: partial fraction (Type: 2)

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