AP EAMCET · Maths · Indefinite Integration
\(\int x \operatorname{Cos}^{-1}\left(\frac{1-x^2}{1+x^2}\right) d x(x>0)=\)
- A \(-x+\left(1+x^2\right) \operatorname{Tan}^{-1} x+c\)
- B \(x-\left(1+x^2\right) \operatorname{Cot}^{-1} x+c\)
- C \(-x+\left(1+x^2\right) \operatorname{Cot}^{-1} x+c\)
- D \(x-\left(1+x^2\right) \operatorname{Tan}^{-1} x+c\)
Answer & Solution
Correct Answer
(A) \(-x+\left(1+x^2\right) \operatorname{Tan}^{-1} x+c\)
Step-by-step Solution
Detailed explanation
Let \(x=\tan\theta\). Then \(\operatorname{Cos}^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \operatorname{Cos}^{-1}(\cos 2\theta) = 2\theta = 2\operatorname{Tan}^{-1}x\). \(\int x (2\operatorname{Tan}^{-1} x) d x = 2 \int x \operatorname{Tan}^{-1} x d x\).…
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