TS EAMCET · Maths · Definite Integration
\(\int_0^{\frac{\pi}{2}} \frac{\sin ^2 x}{\sin x+\cos x} d x=\)
- A \(\sqrt{2} \log (\sqrt{2}+1)\)
- B \(\frac{1}{\sqrt{2}} \log (\sqrt{2}+1)\)
- C \(\log (\sqrt{2}+1)\)
- D \(\frac{1}{\sqrt{2}} \log (\sqrt{2}-1)\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{\sqrt{2}} \log (\sqrt{2}+1)\)
Step-by-step Solution
Detailed explanation
\(I=\int_0^{\pi / 2} \frac{\sin ^2 x}{\sin x+\cos x} d x ... (i)\) \(I=\int_0^{\pi / 2} \frac{\sin ^2\left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)+\cos \left(\frac{\pi}{2}-x\right)} d x\)…
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