CBSE Class 12 Mathematics Indefinite And Definite Integrals Worksheet Set 03

CBSE Class 12 Mathematics Indefinite and Definite Integrals (3). 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: \[ I = \int_{0}^{\pi} \log(1 + \cos x) \, dx \]
Answer:
\[ I = \int_{0}^{\pi} \log(1 + \cos x) \, dx \quad \dots (1) \]
\[ I = \int_{0}^{\pi} \log[1 + \cos(\pi - x)] \, dx \quad \dots \text{(P-IV)} \]
\[ I = \int_{0}^{\pi} \log(1 - \cos x) \, dx \quad \dots (2) \]
Adding (1) and (2):
\[ 2I = \int_{0}^{\pi} \log((1 + \cos x)(1 - \cos x)) \, dx \]
\[ 2I = \int_{0}^{\pi} \log(1 - \cos^2 x) \, dx \]
\[ 2I = \int_{0}^{\pi} \log(\sin^2 x) \, dx \]
\[ 2I = 2 \int_{0}^{\pi} \log(\sin x) \, dx \quad \dots [\log m^n = n \log m] \]
\[ I = \int_{0}^{\pi} \log(\sin x) \, dx \]
\[ I = 2 \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx \quad \dots \text{(P-VI)} \]
\[ \frac{I}{2} = \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx \quad \dots (3) \]
\[ \frac{I}{2} = \int_{0}^{\frac{\pi}{2}} \log\left(\sin\left(\frac{\pi}{2} - x\right)\right) \, dx \quad \dots \text{(P-IV)} \]
\[ \frac{I}{2} = \int_{0}^{\frac{\pi}{2}} \log(\cos x) \, dx \quad \dots (4) \]
Adding (3) and (4):
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\sin x \cdot \cos x) \, dx \]
\[ I = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{\sin(2x)}{2}\right) \, dx \]
\[ I = \int_{0}^{\frac{\pi}{2}} (\log(\sin(2x)) - \log 2) \, dx \]
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\sin(2x)) \, dx - \int_{0}^{\frac{\pi}{2}} \log 2 \, dx \]
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\sin(2x)) \, dx - \log 2 [x]_{0}^{\frac{\pi}{2}} \]
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\sin(2x)) \, dx - \frac{\pi}{2} \log 2 \]
\[ I = I_1 - \frac{\pi}{2} \log 2 \quad \dots (5) \]
Where \( I_1 = \int_{0}^{\frac{\pi}{2}} \log(\sin(2x)) \, dx \)
Put \( 2x = t \)
\( dx = \frac{dt}{2} \)
When \( x = 0 \implies t = 0 \)
When \( x = \frac{\pi}{2} \implies t = \pi \)
\[ \therefore I_1 = \frac{1}{2} \int_{0}^{\pi} \log(\sin t) \, dt \]
\[ I_1 = \frac{1}{2} \times 2 \int_{0}^{\frac{\pi}{2}} \log(\sin t) \, dt \quad \dots \text{(P-VI)} \]
\[ I_1 = \int_{0}^{\frac{\pi}{2}} \log(\sin t) \, dt \]
\[ I_1 = \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx \quad \dots \text{(P-I)} \]
\[ I_1 = \frac{I}{2} \quad \dots \text{\{from eq. (3)\}} \]
\( \therefore \) eq. (5) becomes:
\[ I = \frac{I}{2} - \frac{\pi}{2} \log 2 \]
\[ \Rightarrow I - \frac{I}{2} = -\frac{\pi}{2} \log 2 \]
\[ \Rightarrow \frac{I}{2} = -\frac{\pi}{2} \log 2 \]
\[ I = -\pi \log 2 \quad \text{ans.} \]

Question. Evaluate: \[ I = \int_{0}^{\pi} \frac{x \tan x}{\sec x \cdot \csc x} \, dx \]
Answer:
\[ I = \int_{0}^{\pi} \frac{x \frac{\sin x}{\cos x}}{\frac{1}{\cos x} \cdot \frac{1}{\sin x}} \, dx \]
\[ I = \int_{0}^{\pi} x \sin^2 x \, dx \quad \dots (1) \]
\[ I = \int_{0}^{\pi} (\pi - x) \sin^2(\pi - x) \, dx \quad \dots \text{(P-IV)} \]
\[ I = \int_{0}^{\pi} (\pi - x) \sin^2 x \, dx \quad \dots (2) \]
Adding (1) and (2):
\[ 2I = \int_{0}^{\pi} (x \sin^2 x + \pi \sin^2 x - x \sin^2 x) \, dx \]
\[ 2I = \pi \int_{0}^{\pi} \sin^2 x \, dx \]
\[ 2I = 2\pi \int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx \quad \dots \text{(P-VI)} \]
\[ I = \pi \int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx \]
\[ I = \pi \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos(2x)}{2} \, dx \]
\[ I = \frac{\pi}{2} \left[ x - \frac{\sin(2x)}{2} \right]_{0}^{\frac{\pi}{2}} \]
\[ I = \frac{\pi}{2} \left[ \left( \frac{\pi}{2} - \frac{\sin \pi}{2} \right) - (0 - 0) \right] \]
\[ I = \frac{\pi}{2} \left[ \frac{\pi}{2} - 0 \right] \]
\[ I = \frac{\pi^2}{4} \quad \text{ans.} \]

Question. Evaluate: \[ I = \int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x} + \sqrt{x}} \, dx \]
Answer:
\[ I = \int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x} + \sqrt{x}} \, dx \quad \dots (1) \]
\[ I = \int_{1}^{2} \frac{\sqrt{1+2-x}}{\sqrt{3-(1+2-x)} + \sqrt{1+2-x}} \, dx \quad \dots \left[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \right] \]
\[ I = \int_{1}^{2} \frac{\sqrt{3-x}}{\sqrt{x} + \sqrt{3-x}} \, dx \quad \dots (2) \]
Adding (1) and (2):
\[ 2I = \int_{1}^{2} \frac{\sqrt{x} + \sqrt{3-x}}{\sqrt{3-x} + \sqrt{x}} \, dx \]
\[ 2I = \int_{1}^{2} 1 \cdot dx \]
\[ 2I = [x]_{1}^{2} \]
\[ 2I = 2 - 1 \]
\[ \Rightarrow I = \frac{1}{2} \quad \text{ans.} \]

Question. Evaluate: \[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1 + \sqrt{\cot x}} \, dx \]
Answer:
\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1 + \sqrt{\cot x}} \, dx \quad \dots (1) \]
\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1 + \sqrt{\cot\left(\frac{\pi}{6} + \frac{\pi}{3} - x\right)}} \, dx \quad \dots \text{(P-V) (above)} \]
\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1 + \sqrt{\tan x}} \, dx \]
\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1 + \frac{1}{\sqrt{\cot x}}} \, dx \]
\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + 1} \, dx \quad \dots (2) \]
Adding (1) and (2):
\[ 2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cot x} + 1}{\sqrt{\cot x} + 1} \, dx \]
\[ 2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 \cdot dx \]
\[ 2I = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} \]
\[ 2I = \frac{\pi}{3} - \frac{\pi}{6} \]
\[ 2I = \frac{\pi}{6} \]
\[ I = \frac{\pi}{12} \quad \text{ans.} \]

Question. Evaluate: \[ I = \int_{0}^{1} |5x - 3| \, dx \]
Answer:
\[ I = -\int_{0}^{\frac{3}{5}} (5x - 3) \, dx + \int_{\frac{3}{5}}^{1} (5x - 3) \, dx \]
\[ I = -\left[ \frac{5x^2}{2} - 3x \right]_{0}^{\frac{3}{5}} + \left[ \frac{5x^2}{2} - 3x \right]_{\frac{3}{5}}^{1} \]
\[ I = -\left[ \frac{5}{2} \left(\frac{9}{25}\right) - 3\left(\frac{3}{5}\right) \right] + \left[ \left(\frac{5}{2} - 3\right) - \left(\frac{5}{2} \left(\frac{9}{25}\right) - 3\left(\frac{3}{5}\right)\right) \right] \]
\[ I = -\left( \frac{9}{10} - \frac{9}{5} \right) + \left[ -\frac{1}{2} - \left(\frac{9}{10} - \frac{9}{5}\right) \right] \]
\[ I = -\frac{9}{10} + \frac{9}{5} - \frac{1}{2} - \frac{9}{10} + \frac{9}{5} \]
\[ I = \frac{13}{10} \quad \text{ans.} \]

Question. Evaluate: \[ I = \int_{-5}^{5} |x - 2| \, dx \]
Answer:
\[ I = -\int_{-5}^{2} (x - 2) \, dx + \int_{2}^{5} (x - 2) \, dx \]
\[ I = -\left[ \frac{x^2}{2} - 2x \right]_{-5}^{2} + \left[ \frac{x^2}{2} - 2x \right]_{2}^{5} \]
\[ I = -\left[ (2 - 4) - \left(\frac{25}{2} + 10\right) \right] + \left[ \left(\frac{25}{2} - 10\right) - (2 - 4) \right] \]
\[ I = -\left[ -2 - \frac{45}{2} \right] + \left[ \frac{5}{2} + 2 \right] \]
\[ I = 2 + \frac{45}{2} + \frac{5}{2} + 2 = 29 \quad \text{ans.} \]

Question. Evaluate: \[ I = \int_{1}^{5} |x - 6| \, dx \]
Answer:
\[ I = -\int_{1}^{5} (x - 6) \, dx \]
\[ I = -\left[ \frac{x^2}{2} - 6x \right]_{1}^{5} \]
\[ I = -\left[ \left(\frac{25}{2} - 30\right) - \left(\frac{1}{2} - 6\right) \right] \]
\[ I = -\left[ -\frac{35}{2} + \frac{11}{2} \right] = 12 \quad \text{ans.} \]

Question. Evaluate: \[ I = \int_{-1}^{1} e^{|x|} \, dx \]
Answer:
\[ I = \int_{-1}^{0} e^{-x} \, dx + \int_{0}^{1} e^x \, dx \]
\[ I = \left[ \frac{e^{-x}}{-1} \right]_{-1}^{0} + \left[ e^x \right]_{0}^{1} \]
\[ I = \left[ \frac{1}{-1} - \frac{e^1}{-1} \right] + [e^1 - e^0] \]
\[ I = [-1 + e] + [e - 1] \]
\[ I = 2e - 2 \quad \text{ans.} \]

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

Question. Evaluate: \[ I = \int_{0}^{2} |x^2 + 2x - 3| \, dx \]
Answer:
\[ I = \int_{0}^{2} |(x+3)(x-1)| \, dx \]
\[ \therefore I = -\int_{0}^{1} (x^2 + 2x - 3) \, dx + \int_{1}^{2} (x^2 + 2x - 3) \, dx \]
\[ I = -\left[ \frac{x^3}{3} + x^2 - 3x \right]_{0}^{1} + \left[ \frac{x^3}{3} + x^2 - 3x \right]_{1}^{2} \]
\[ I = -\left[ \left( \frac{1}{3} + 1 - 3 \right) - (0) \right] + \left[ \left( \frac{8}{3} + 4 - 6 \right) - \left( \frac{1}{3} + 1 - 3 \right) \right] \]
\[ I = -\left[ -\frac{5}{3} \right] + \left[ \frac{2}{3} + \frac{5}{3} \right] \]
\[ I = \frac{12}{3} = 4 \quad \text{ans.} \]

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