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JEE Mains · Maths · STD 12 - 7.1 indefinite integral

For \(\mathrm{x} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), if \(y(x)=\int \frac{\operatorname{cosec} x+\sin x}{\operatorname{cosec} x \sec x+\tan x \sin ^2 x} d x\) and \(\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}} y(x)=0\) then \(y\left(\frac{\pi}{4}\right)\) is equal to

  1. A \(\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
  2. B \(\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
  3. C  \(-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
  4. D \(\frac{1}{\sqrt{2}} \tan ^{-1}\left(-\frac{1}{2}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{1}{\sqrt{2}} \tan ^{-1}\left(-\frac{1}{2}\right)\)

Step-by-step Solution

Detailed explanation

\(y(x)=\int \frac{\left(1+\sin ^2 x\right) \cos x}{1+\sin ^4 x} d x\) Put \(\sin x=t\) \( =\int \frac{1+t^2}{t^4+1} d t=\frac{1}{\sqrt{2}} \tan ^{-1} \frac{\left(t-\frac{1}{t}\right)}{\sqrt{2}}+C \) \( x=\frac{\pi}{2}, t=1 \quad \therefore C=0 \)…
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