SATHEE: Indefinite Integration 1 Question 81

Indefinite Integration 1 Question 81

82. Evaluate $\int_{0}^{1} \log [\sqrt{(1 - x)} + \sqrt{(1 + x)}] d x$ .

(1988, 5M)

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Answer:

Correct Answer: 82. $\frac{1}{2} \log 2 - \frac{1}{2} + \frac{π}{4}$

Solution:

Let $I = \int_{0}^{1} \log (\sqrt{1 - x} + \sqrt{1 + x}) d x$

Put $x = \cos 2 θ$

$\Rightarrow d x = - 2 \sin 2 θ d θ$

$∴ I = - 2 \int_{π / 4}^{0} \log [\sqrt{1 - \cos 2 θ} + \sqrt{1 + \cos 2 θ}] (\sin 2 θ) d θ$

$= - 2 \int_{π / 4}^{0} \log [\sqrt{2} (\sin θ + \cos θ)] \sin 2 θ d θ$

$= - 2 \int_{π / 4}^{0} [(\log \sqrt{2}) \sin 2 θ$

$+ \log (\sin θ + \cos θ) \cdot \sin 2 θ] d θ$

$= - 2 \log \sqrt{2} {\frac{- \cos 2 θ^{0}}{2}}_{π / 4}^{0}$

$- 2 \int_{π / 4}^{0} \log (\sin θ + \cos θ) \cdot \sin 2 θ d θ$

$= \log \sqrt{2} - 2 - \log (\sin θ + \cos θ) \cdot {\frac{\cos 2 θ}{2}}_{π / 4}^{0}$

$- \int_{π / 4}^{0} \frac{\cos θ - \sin θ}{\cos θ + \sin θ} \times \frac{- \cos 2 θ}{2} d θ$

$= \log (\sqrt{2}) - 20 + \frac{1}{2} \int_{π / 4}^{0} (\cos θ - \sin θ)^{2} d θ$

$= \frac{1}{2} \log 2 - \int_{π / 4}^{0} (1 - \sin 2 θ) d θ$

$= \frac{1}{2} \log 2 - θ + {\frac{\cos 2 θ}{2}}_{π / 4}^{0}$

$= \frac{1}{2} \log 2 - \frac{1}{2} - \frac{π}{4} = \frac{1}{2} \log 2 - \frac{1}{2} + \frac{π}{4}$

Indefinite Integration 1 Question 81

82. Evaluate $\int_{0}^{1} \log [\sqrt{(1 - x)} + \sqrt{(1 + x)}] d x$ .

Answer:

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Indefinite Integration 1 Question 81

82. Evaluate ∫01log⁡[(1−x)+(1+x)]dx.

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82. Evaluate $\int_{0}^{1} \log [\sqrt{(1 - x)} + \sqrt{(1 + x)}] d x$ .