Question

\(I(x) = \int \frac{x^2(xsec^2x + tanx)}{(xtanx+1)^2} dx\). If \(I(0)=1\) then \(I(\frac{\pi}{4})\) is equal to

• A. \(-\frac{\pi^2}{4\pi+16}+2ln(\frac{\pi+4}{4\sqrt2})+1\)
• B. \(\frac{\pi^2}{4\pi+16}-2ln(\frac{\pi+4}{4\sqrt2})+1\)
• C. \(-\frac{\pi^2}{\pi+4}+2ln(\frac{\pi+1}{\sqrt2})+1\)
• D. \(\frac{\pi^2}{\pi+16}+2ln(\frac{\pi+1}{4\sqrt2})+1\)

Answer 1

The Correct Option is A

Using integration by parts
\(I(x) = x^2.\frac{(-1)}{x\:tanx+1}\:-\:\int\:2x.\frac{(-1)}{x\:tanx+1}dx\)
\(=\,-\frac{x^2}{x\:tanx+1}+2\int\frac{x\:cos\,x}{x\:sin\,x+cos\,x}dx\)
\(\Rightarrow I(x)=-\frac{x^2}{x\:tan\,x+1}+2ln|x\:sin\,x+cos\,x|+c\)
put \(x=0\)
\(c=1\)
\(\therefore I(\frac{\pi}{4})=\frac{\frac{-\pi^2}{16}}{\frac{\pi}{4}+1}+2ln(\frac{\frac{\pi}{4}+1}{\sqrt{2}})+1\)
\(I(\frac{\pi}{4})= -\frac{\pi^2}{4\pi+16}+2ln(\frac{\pi+4}{4\sqrt{2}})+1\)

So, the correct answer is (A): \(-\frac{\pi^2}{4\pi+16}+2ln(\frac{\pi+4}{4\sqrt2})+1\)