Question:
The value of integral \(\displaystyle\int_{\pi/4}^{3\pi/4} \dfrac{x}{1+\sin x}\,dx\) is :
The value of integral \(\displaystyle\int_{\pi/4}^{3\pi/4} \dfrac{x}{1+\sin x}\,dx\) is :
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Replace x by pi - x, which maps the interval pi/4 to 3pi/4 onto itself, and add the two forms.
Updated On: Nov 11, 2026
- \(\pi\sqrt2\)
- \(\pi\left(\sqrt2 - 1\right)\)
- \(\dfrac\pi2\left(\sqrt2 + 1\right)\)
- \(2\pi\left(\sqrt2 - 1\right)\)
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The Correct Option is B
Solution and Explanation
Let \(I = \int_{\pi/4}^{3\pi/4}\dfrac{x}{1+\sin x}dx\). Replace \(x\) by \(\pi - x\) (the interval maps onto itself and \(\sin(\pi-x)=\sin x\)): \(I = \int_{\pi/4}^{3\pi/4}\dfrac{\pi - x}{1+\sin x}dx\).
Adding: \(2I = \pi\int_{\pi/4}^{3\pi/4}\dfrac{dx}{1+\sin x}\).
\(\dfrac{1}{1+\sin x} = \dfrac{1-\sin x}{\cos^2x} = \sec^2x - \sec x\tan x\), with antiderivative \(\tan x - \sec x\).
At \(3\pi/4\): \(-1 + \sqrt2\). At \(\pi/4\): \(1 - \sqrt2\). Difference \(= 2(\sqrt2-1)\).
\(2I = 2\pi(\sqrt2-1)\), so \(I = \pi(\sqrt2-1)\).
Answer: (2) \(\pi\left(\sqrt2 - 1\right)\)
Adding: \(2I = \pi\int_{\pi/4}^{3\pi/4}\dfrac{dx}{1+\sin x}\).
\(\dfrac{1}{1+\sin x} = \dfrac{1-\sin x}{\cos^2x} = \sec^2x - \sec x\tan x\), with antiderivative \(\tan x - \sec x\).
At \(3\pi/4\): \(-1 + \sqrt2\). At \(\pi/4\): \(1 - \sqrt2\). Difference \(= 2(\sqrt2-1)\).
\(2I = 2\pi(\sqrt2-1)\), so \(I = \pi(\sqrt2-1)\).
Answer: (2) \(\pi\left(\sqrt2 - 1\right)\)
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