For x ∈ (−π/2, π/2), if y(x) = ∫ cosecx + sinx / cosecx secx + tanx sin^2x dx and

Question

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

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

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

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

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

Answer 1

Correct option is (4) \(\frac{1}{\sqrt{2}} \tan ^{-1}\left(-\frac{1}{2}\right)\) 

\(\mathrm {y(x)=\int \frac{\left(1+\sin ^{2} x\right) \cos x}{1+\sin ^{4} x} d x}\)

Put \(\sin \mathrm{x}=\mathrm{t}\)

\(=\int \frac{1+\mathrm{t}^{2}}{\mathrm{t}^{4}+1} \mathrm{dt}=\frac{1}{\sqrt{2}} \tan ^{-1} \frac{\left(\mathrm{t}-\frac{1}{\mathrm{t}}\right)}{\sqrt{2}}+\mathrm{C}\)

\(\mathrm{x}=\frac{\pi}{2}, \mathrm{t}=1 \quad \therefore \mathrm{C}=0\)

\(\mathrm y\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \tan ^{-1}\left(-\frac{1}{2}\right)\)