Indefinite Integration 1 Question 60
61. The value of $\int _{\pi / 4}^{3 \pi / 4} \frac{x}{1+\sin x} d x \ldots \ldots$.
$(1993,2 M)$
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Solution:
- Let $I=\int _{\pi / 4}^{3 \pi / 4} \frac{x}{1+\sin x} d x$
$$ \begin{aligned} & \Rightarrow I=\int _{\pi / 4}^{3 \pi / 4} \frac{\frac{\pi}{4}+\frac{3 \pi}{4}-x}{1+\sin \frac{\pi}{4}+\frac{3 \pi}{4}-x} d x \\ & {\left[\because \int _a^{b} f(x) d x=\int _a^{b} f(a+b-x) d x\right] } \end{aligned} $$
$=\int _{\pi / 4}^{3 \pi / 4} \frac{\pi-x}{1+\sin (\pi-x)} d x$
$=\int _{\pi / 4}^{3 \pi / 4} \frac{\pi}{1+\sin x} d x-\int _{\pi / 4}^{3 \pi / 4} \frac{x}{1+\sin x} d x$
$=\pi \int _{\pi / 4}^{3 \pi / 4} \frac{d x}{1+\sin x}-I$
[from Eq. (i)]
$=\frac{\pi}{2} \int _{\pi / 4}^{3 \pi / 4} \frac{d x}{(1+\sin x)}$
$=\frac{\pi}{2} \int _{\pi / 4}^{3 \pi / 4} \frac{(1-\sin x)}{(1+\sin x)(1-\sin x)} d x$
$=\frac{\pi}{2} \int _{\pi / 4}^{3 \pi / 4} \frac{(1-\sin x)}{1-\sin ^{2} x} d x$
$=\frac{\pi}{2} \int _{\pi / 4}^{3 \pi / 4} \frac{1}{\cos ^{2} x}-\frac{\sin x}{\cos ^{2} x} d x$
$=\frac{\pi}{2} \int _{\pi / 4}^{3 \pi / 4}\left(\sec ^{2} x-\sec x \cdot \tan x\right) d x$
$=\frac{\pi}{2}[\tan x-\sec x] _{\pi / 4}^{3 \pi / 4}$
$=\frac{\pi}{2}[-1-1-(-\sqrt{2}-\sqrt{2})]$
$=\frac{\pi}{2}(-2+2 \sqrt{2})=\pi(\sqrt{2}-1)$
