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JEE Mains · Maths · STD 12 - 7.2 definite integral
The value of integral \(\int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{x}{{1 + \sin x}}} dx\) is
- A \(\frac{\pi }{2}\left( {\sqrt 2 + 1} \right)\)
- B \(\pi \left( {\sqrt 2 - 1} \right)\)
- C \(2\pi \left( {\sqrt 2 - 1} \right)\)
- D \(\pi \sqrt 2 \)
Answer & Solution
Correct Answer
(A) \(\frac{\pi }{2}\left( {\sqrt 2 + 1} \right)\)
Step-by-step Solution
Detailed explanation
Let \(I=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{x}{1+\sin x} d x\) also let \(K=\frac{x}{1+\sin x}\) Multiplying numerator and denominator by \((1-\sin x),\) we get \(K=\frac{x(1-\sin x)}{1-(\sin x)^{2}}\) \(=\frac{x(1-\sin x)}{(\cos x)^{2}}\) \(=x(1-\sin x) \sec ^{2} x\)…
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