AP EAMCET · Maths · Definite Integration
\(\int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x=\)
- A \(\pi-\log 2\)
- B \(\pi+\log 2\)
- C \(\frac{\pi}{2}-\log 2\)
- D \(\frac{\pi}{2}+\log 2\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{2}-\log 2\)
Step-by-step Solution
Detailed explanation
\(I=\int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x\) When, \(x=\tan \theta \text {, }\) \(\frac{2 x}{1+x^2}=\frac{2 \tan \theta}{1+\tan ^2 \theta}=\sin 2 \theta\) Also, \(\quad d x=\sec ^2 \theta d \theta\) and, When \(x=0, \theta=0\) When \(x=1, \theta=\frac{\pi}{4}\)…
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