CBSE Class 12 Mathematics Indefinite And Definite Integrals Worksheet Set 06

CBSE Class 12 Mathematics Indefinite and Definite Integrals (6). 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: \[ I = \int_{3}^{4} e^{3x} + 2x \, dx \]
Answer:
\( a = 3, b = 4, nh = 4 - 3 = 1 \)
\[ I = \lim_{h \to 0} h [f(3) + f(3+h) + f(3+2h) + \dots + f(3+(n-1)h)] \]
\[ = \lim_{h \to 0} h [(e^9 + 6) + (e^{9+3h} + 6 + 2h) + (e^{9+6h} + 6 + 4h) + \dots + (e^{9+3(n-1)h} + 6 + 2(n-1)h)] \]
\[ = \lim_{h \to 0} h [\{e^9 + e^{9+3h} + e^{9+6h} + \dots + e^{9+3(n-1)h}\} + \{(6 + 6 + 6 + \dots n \text{ term}) + (2h + 4h + \dots 2(n-1)h)\}] \]
\[ = \lim_{h \to 0} h [e^9 + e^{9+3h} + e^{9+6h} + \dots + e^{9+3(n-1)h}] + \lim_{h \to 0} h [6n + 2h(1 + 2 + \dots + (n-1))] \]
\[ = e^9 \lim_{h \to 0} h [1 + e^{3h} + e^{6h} + \dots + e^{3(n-1)h}] + \lim_{h \to 0} h \left[6n + \frac{2h \cdot n(n-1)}{2}\right] \]
\[ = e^9 \lim_{h \to 0} h \left[1 \left(\frac{(e^{3h})^n - 1}{e^{3h}-1}\right)\right] + \lim_{h \to 0} [6nh + (nh)(nh - h)] \]
\[ = e^9 \lim_{h \to 0} h \left[\frac{e^{3nh} - 1}{\frac{e^{3h}-1}{3h} \times 3h}\right] + \lim_{h \to 0} [6nh + (nh)(nh - h)] \]

Put \( nh = 1 \):
\[ \therefore I = e^9 \frac{(e^3 - 1)}{\lim_{h \to 0} \left(\frac{e^{3h}-1}{3h}\right) \times 3} + \lim_{h \to 0} [6 + (1)(1 - h)] \]
\[ I = \frac{e^9(e^3 - 1)}{3} + 6 + 1 \]
\[ I = \frac{e^9(e^3 - 1)}{3} + 7 \quad \text{Ans.} \]

Question. Evaluate: \[ I = \int_{\frac{\pi}{2}}^{\pi} e^x \left(\frac{1-\sin x}{1-\cos x}\right) \, dx \]
Answer:
\[ I = \int_{\frac{\pi}{2}}^{\pi} e^x \left(\frac{1-\sin x}{1-\cos x}\right) \, dx \target{} \]
\[ I = \int_{\frac{\pi}{2}}^{\pi} e^x \left(\frac{1-2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}}\right) \, dx \]
\[ I = \int_{\frac{\pi}{2}}^{\pi} e^x \left(\frac{1}{2} \csc^2 \left(\frac{x}{2}\right) - \cot \left(\frac{x}{2}\right)\right) \, dx \]
\[ I = \int_{\frac{\pi}{2}}^{\pi} e^x \cot \left(\frac{x}{2}\right) \, dx + \frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} e^x \cdot \csc^2 \left(\frac{x}{2}\right) \, dx \]
\[ I = - \left[ \left(\cot \frac{x}{2} \cdot e^x\right)_{\pi/2}^{\pi} - \int_{\pi/2}^{\pi} -\frac{1}{2} \cdot \csc^2 \left(\frac{x}{2}\right) e^x \, dx \right] + \frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} e^x \csc^2 \left(\frac{x}{2}\right) \, dx \]
\[ I = - \left[ \cot \frac{\pi}{2} \cdot e^{\pi} - \cot \frac{\pi}{4} \cdot e^{\pi/2} \right] - \frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} e^x \csc^2 \left(\frac{x}{2}\right) \, dx + \frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} e^x \csc^2 \left(\frac{x}{2}\right) \, dx \]
\[ I = - \left[0 - e^{\pi/2}\right] \quad \dots \{\because \cot \left(\frac{\pi}{2}\right) = 0, \cot \left(\frac{\pi}{4}\right) = 1\} \]
\[ \therefore I = e^{\frac{\pi}{2}} \quad \text{Ans.} \]

Question. Evaluate: \[ I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \cdot \tan^{-1}(\sin x) \, dx \]
Answer:
\[ I = 2 \int_{0}^{\frac{\pi}{2}} \sin x \cdot \cos x \cdot \tan^{-1}(\sin x) \, dx \]
Put \( \sin x = t \), when \( x = 0 \Rightarrow t = 0 \)
\( \cos x \, dx = dt \), when \( x = \frac{\pi}{2} \Rightarrow t = 1 \)
\[ \therefore I = 2 \int_{0}^{1} t \cdot \tan^{-1}(t) \, dt \]
\[ = 2 \left[ \left( \tan^{-1} t \cdot \frac{t^2}{2} \right)_{0}^{1} - \int_{0}^{1} \frac{1}{1+t^2} \cdot \frac{t^2}{2} \, dt \right] \]
\[ = 2 \left[ \left(\frac{\pi}{4} \cdot \frac{1}{2}\right) - 0 - \frac{1}{2} \int_{0}^{1} \frac{t^2}{1+t^2} \, dt \right] \]
\[ = 2 \left[ \frac{\pi}{8} - \frac{1}{2} \int_{0}^{1} \frac{1+t^2-1}{1+t^2} \, dt \right] \]
\[ = \frac{\pi}{4} - \int_{0}^{1} 1 - \frac{1}{1+t^2} \, dt \]
\[ = \frac{\pi}{4} - [t - \tan^{-1} t]_{0}^{1} \]
\[ = \frac{\pi}{4} - \left[\left(1 - \frac{\pi}{4}\right) - 0\right] \]
\[ I = \frac{\pi}{2} - 1 \quad \text{Ans.} \]

Question. Evaluate: \[ I = \int_{0}^{\frac{\pi}{2}} \frac{x+\sin x}{1+\cos x} \, dx \]
Answer:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{x+2 \sin\left(\frac{x}{2}\right) \cdot \cos\left(\frac{x}{2}\right)}{2 \cos^2\left(\frac{x}{2}\right)} \, dx \]
\[ I = \int_{0}^{\frac{\pi}{2}} \left( \frac{x}{2\cos^2(x/2)} + \frac{2 \sin\left(\frac{x}{2}\right) \cdot \cos\left(\frac{x}{2}\right)}{2 \cos^2\left(\frac{x}{2}\right)} \right) \, dx \]
\[ I = \int_{0}^{\frac{\pi}{2}} \left( \frac{1}{2} x \sec^2 \left(\frac{x}{2}\right) + \tan \left(\frac{x}{2}\right) \right) \, dx \]
\[ I = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} x \cdot \sec^2 \left(\frac{x}{2}\right) \, dx + \int_{0}^{\frac{\pi}{2}} \tan \left(\frac{x}{2}\right) \, dx \]
\[ I = \frac{1}{2} \left[ \left(x \cdot \tan \left(\frac{x}{2}\right) \cdot 2\right)_{0}^{\pi/2} - \int_{0}^{\frac{\pi}{2}} 1 \cdot \tan \left(\frac{x}{2}\right) \cdot 2 \, dx \right] + \int_{0}^{\frac{\pi}{2}} \tan \left(\frac{x}{2}\right) \, dx \]
\[ I = \frac{1}{2} \left[ \left(\frac{\pi}{2} \cdot \tan \frac{\pi}{4} \cdot 2\right) - 0 \right] - \int_{0}^{\frac{\pi}{2}} \tan \frac{x}{2} \, dx + \int_{0}^{\frac{\pi}{2}} \tan \frac{x}{2} \, dx \]
\[ I = \frac{1}{2} \left[\frac{\pi}{1}\right] = \frac{\pi}{2} \quad \text{Ans.} \]

Question. Evaluate: \[ I = \int_{0}^{\frac{\pi}{4}} \sin^3(2t) \cdot \cos(2t) \, dt \]
Answer:
Put \( \sin(2t) = z \)
\( \therefore 2 \cos(2t) \, dt = dz \), when \( t = 0 \Rightarrow z = 0 \)
\( \cos(2t) \, dt = \frac{dz}{2} \), when \( t = \frac{\pi}{4} \Rightarrow z = 1 \)
\[ \therefore I = \frac{1}{2} \int_{0}^{1} z^3 \, dz \]
\[ = \frac{1}{2} \left(\frac{z^4}{4}\right)_{0}^{1} \]
\[ = \frac{1}{2} \left(\frac{1}{4} - 0\right) \]
\[ I = \frac{1}{8} \quad \text{Ans.} \target{} \]

Question. Evaluate: \[ I = \int_{4}^{9} \frac{\sqrt{x}}{\left(30-x^{\frac{3}{2}}\right)^2} \, dx \]
Answer:
Put \( 30 - x^{3/2} = t \), when \( x = 4 \Rightarrow t = 30 - 8 = 22 \)
\( -\frac{3}{2} x^{1/2} \, dx = dt \), when \( x = 9 \Rightarrow t = 30 - 27 = 3 \)
\( \sqrt{x} \, dx = \frac{-2}{3} \, dt \)
\[ \therefore I = \frac{-2}{3} \int_{22}^{3} \frac{dt}{t^2} \]
\[ = \frac{-2}{3} \left(-\frac{1}{t}\right)_{22}^{3} \]
\[ = \frac{2}{3} \left[\frac{1}{3} - \frac{1}{22}\right] \]
\[ = \frac{2}{3} \left( \frac{22-3}{66} \right) = \frac{2(19)}{3 \times 66} = \frac{19}{99} \quad \text{Ans.} \]

Question. Evaluate: \[ \int_{0}^{\frac{\pi}{2}} \sqrt{\sin\varphi} \cdot \cos^5\varphi \, d\varphi \]
Answer:
\[ I = \int_{0}^{\frac{\pi}{2}} \sqrt{\sin\varphi} \cdot \cos^4\varphi \cdot \cos\varphi \, d\varphi \]
Put \( \sin\varphi = t \), when \( \varphi = 0 \Rightarrow t = 0 \)
\( \cos\varphi \, d\varphi = dt \), when \( \varphi = \frac{\pi}{2} \Rightarrow t = 1 \)
\[ \therefore I = \int_{0}^{1} \sqrt{t} (1 - \sin^2\varphi)^2 \cdot dt \]
\[ = \int_{0}^{1} \sqrt{t}(1 - t^2)^2 \, dt \]
\[ = \int_{0}^{1} \sqrt{t}(1 + t^4 - 2t^2) \, dt \]
\[ = \int_{0}^{1} \left( \sqrt{t} + t^{9/2} - 2t^{5/2} \right) \, dt \]
\[ = \left( \frac{2}{3} t^{3/2} + \frac{2}{11} t^{11/2} - 2 \cdot \frac{2}{7} t^{7/2} \right)_{0}^{1} \]
\[ = \left( \frac{2}{3} + \frac{2}{11} - \frac{4}{7} \right) - (0) \]
\[ = \frac{154+42-132}{231} \]
\[ = \frac{64}{231} \quad \text{Ans.} \]

Question. Evaluate: \[ I = \int_{0}^{1} \sin^{-1} \left(\frac{2x}{1+x^2}\right) \, dx \]
Answer:
\[ I = \int_{0}^{1} \sin^{-1} \left(\frac{2x}{1+x^2}\right) \, dx \]
\[ I = 2 \int_{0}^{1} \tan^{-1}x \, dx \quad \dots \left\{\because 2\tan^{-1}x = \sin^{-1} \left(\frac{2x}{1+x^2}\right)\right\} \]
\[ I = 2 \int_{0}^{1} \tan^{-1} x \cdot 1 \, dx \]
\[ = 2 \left[ (\tan^{-1}x \cdot x)_{0}^{1} - \int_{0}^{1} \frac{1}{1+x^2} \cdot x \, dx \right] \]

Put \( 1 + x^2 = t \), when \( x = 0 \Rightarrow t = 1 \)
\( x \, dx = \frac{dt}{2} \), when \( x = 1 \Rightarrow t = 2 \)

\[ \therefore I = 2 \left[ \left(\frac{\pi}{4} \cdot 1\right) - (0) - \frac{1}{2} \int_{1}^{2} \frac{dt}{t} \right] \]
\[ = 2 \left[ \frac{\pi}{4} - \frac{1}{2} (\log t)_{1}^{2} \right] \]
\[ = 2 \left[ \frac{\pi}{4} - \frac{1}{2} (\log 2 - \log 1) \right] \]
\[ = 2 \left[ \frac{\pi}{4} - \frac{1}{2} \log 2 \right] \quad \dots \{\because \log 1 = 0\} \]
\[ I = \frac{\pi}{2} - \log 2 \quad \text{Ans.} \]

Question. Evaluate: \[ I = \int_{0}^{1} \frac{1}{\sqrt{1+x}-\sqrt{x}} \, dx \]
Answer:
Rationalize:
\[ I = \int_{0}^{1} \frac{\sqrt{1+x}+\sqrt{x}}{1+x-x} \, dx \]
\[ I = \left[ \frac{2}{3} (1 + x)^{3/2} + \frac{2}{3} x^{3/2} \right]_{0}^{1} \]
\[ = \frac{2}{3} \left[ (2)^{3/2} + (1)^{3/2} \right] - \frac{2}{3} \left[ (1)^{3/2} + 0 \right] \]
\[ = \frac{2}{3} (2\sqrt{2} + 1) - \frac{2}{3} (1) \]
\[ = \frac{4\sqrt{2}}{3} + \frac{2}{3} - \frac{2}{3} \]
\[ I = \frac{4\sqrt{2}}{3} \quad \text{ans.} \]

Question. Evaluate: \[ I = \int_{0}^{1} \frac{2x+3}{5x^2+1} \, dx \]
Answer:
\[ I = \int_{0}^{1} \frac{2x+3}{5x^2+1} \, dx \quad (\text{separate}) \]
\[ I = 2 \int_{0}^{1} \frac{x}{5x^2+1} \, dx + 3 \int_{0}^{1} \frac{1}{5x^2+1} \, dx \]

Put \( 5x^2 + 1 = t \), when \( x = 0 \Rightarrow t = 1 \)
\( 10x \, dx = dt \), when \( x = 1 \Rightarrow t = 6 \)
\( x \, dx = \frac{dt}{10} \)

\[ \therefore I = \frac{2}{10} \int_{1}^{6} \frac{dt}{t} + \frac{3}{5} \int_{0}^{1} \frac{1}{x^2+\left(\frac{1}{\sqrt{5}}\right)^2} \, dt \]
\[ = \frac{1}{5} [\log t]_{1}^{6} + \frac{3}{5} \times \sqrt{5} \left[ \tan^{-1}(x\sqrt{5}) \right]_{0}^{1} \]
\[ = \frac{1}{5} \left[ (\log 6 - \log 1) + \frac{3}{\sqrt{5}} (\tan^{-1}\sqrt{5} - \tan^{-1}0) \right] \]
\[ I = \frac{1}{5} \log 6 + \frac{3}{\sqrt{5}} \tan^{-1}(\sqrt{5}) \quad \text{ans.} \]

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