AP EAMCET · Maths · Definite Integration
\(\int_0^{\frac{\pi}{2}} \frac{\sin ^2 x}{\sin x+\cos x} d x=\)
- A \(\frac{3}{\sqrt{2}} \log (\sqrt{2}+1)^{\frac{1}{2}}\)
- B \(\frac{1}{\sqrt{2}} \log (\sqrt{2}+1)\)
- C \(\frac{\sqrt{2}}{3} \log (\sqrt{3}+1)\)
- D \(\frac{2}{\sqrt{3}} \log (\sqrt{2}-1)\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{\sqrt{2}} \log (\sqrt{2}+1)\)
Step-by-step Solution
Detailed explanation
Let \(I=\int_0^{\frac{\pi}{2}} \frac{\sin ^2 x}{\sin x+\cos x} d x\) ...(i) On applying property \(\int_0^a f(x) d x=\int_0^a f(a-x) d x\), we get, \(I=\int_0^{\frac{\pi}{2}} \frac{\cos ^2 x}{\cos x+\sin x} d x\) ...(ii) On adding Eqs. (i) and (ii), we get…
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