Definite Integration Question 80
Question 80
- Evaluate $\int_{0}^{1} \log [\sqrt{(1-x)}+\sqrt{(1+x)}] d x$.
(1988, 5M)
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Answer:
Correct Answer: 82. $\frac{1}{2} \log 2-\frac{1}{2}+\frac{\pi}{4}$
Solution:
- Let $I=\int_{0}^{1} \log (\sqrt{1-x}+\sqrt{1+x}) d x$
Put $x=\cos 2 \theta$
$\Rightarrow d x=-2 \sin 2 \theta d \theta$
$\therefore I=-2 \int_{\pi / 4}^{0} \log [\sqrt{1-\cos 2 \theta}+\sqrt{1+\cos 2 \theta}](\sin 2 \theta) d \theta$
$=-2 \int_{\pi / 4}^{0} \log [\sqrt{2}(\sin \theta+\cos \theta)] \sin 2 \theta d \theta$
$=-2 \int_{\pi / 4}^{0}[(\log \sqrt{2}) \sin 2 \theta$
$+\log (\sin \theta+\cos \theta) \cdot \sin 2 \theta] d \theta$
$=-2 \log \sqrt{2}{\frac{-\cos 2 \theta^{0}}{2}}_{\pi / 4}^{0}$
$-2 \int_{\pi / 4}^{0} \log (\sin \theta+\cos \theta) \cdot \sin 2 \theta d \theta$
$=\log \sqrt{2}-2-\log (\sin \theta+\cos \theta) \cdot \frac{\cos 2 \theta}{2}_{\pi / 4}{ }^{0}$
$-\int_{\pi / 4}^{0} \frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta} \times \frac{-\cos 2 \theta}{2} d \theta$
$=\log (\sqrt{2})-20+\frac{1}{2} \int_{\pi / 4}^{0}(\cos \theta-\sin \theta)^{2} d \theta$
$=\frac{1}{2} \log 2-\int_{\pi / 4}^{0}(1-\sin 2 \theta) d \theta$
$=\frac{1}{2} \log 2-\theta+\frac{\cos 2 \theta}{2}_{\pi / 4}^{0}$
$=\frac{1}{2} \log 2-\frac{1}{2}-\frac{\pi}{4}=\frac{1}{2} \log 2-\frac{1}{2}+\frac{\pi}{4}$
