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): \(-\frac{\pi^2}{4\pi+16}+2ln(\frac{\pi+4}{4\sqrt2})+1\)
 

Answer 2

Integral Calculation 

We are given $I(x) = \int \frac{x^2 (x \sec^2 x + \tan x)}{(x \tan x + 1)^2} dx$.

We can rewrite the integrand as:

$\frac{x^2 (x \sec^2 x + \tan x)}{(x \tan x + 1)^2} = \frac{x^3 \sec^2 x + x^2 \tan x}{(x \tan x + 1)^2}$

Let $u = x^2$ and $dv = \frac{x \sec^2 x + \tan x}{(x \tan x + 1)^2} dx$. If $v = \frac{-1}{x \tan x + 1}$, then $dv = \frac{x \sec^2 x + \tan x}{(x \tan x + 1)^2} dx$. This can be verified by differentiating $v$ with respect to $x$. Now, using integration by parts:

$\int u dv = uv - \int v du$

$I(x) = x^2 \left( \frac{-1}{x \tan x + 1} \right) - \int \left( \frac{-1}{x \tan x + 1} \right) (2x) dx$

$I(x) = \frac{-x^2}{x \tan x + 1} + \int \frac{2x}{x \tan x + 1} dx$

We know that $\int \frac{f'(x)}{f(x)} dx = \ln |f(x)| + C$. The derivative of $x \tan x + 1 $ is $x \sec^2 x + \tan x$. Therefore,

$\int \frac{2x}{x \tan x + 1} dx = 2 \ln |x \tan x + 1 | + C$

Thus,

$I(x) = \frac{-x^2}{x \tan x + 1} + 2 \ln |x \tan x + 1| + C$

Since I(0) = 0, we have 0 = 0 + 2 ln(1) + C ⇒ C = 0. Therefore,

$I(x) = \frac{-x^2}{x \tan x + 1} + 2 \ln |x \tan x + 1|$

Now, we need to find $I(\frac{\pi}{4})$:

$I(\frac{\pi}{4}) = \frac{-(\frac{\pi}{4})^2}{\frac{\pi}{4} \tan(\frac{\pi}{4}) + 1} + 2 \ln |\frac{\pi}{4} \tan(\frac{\pi}{4}) + 1| = \frac{-\frac{\pi^2}{16}}{\frac{\pi}{4} + 1} + 2 \ln |\frac{\pi}{4} + 1| = \frac{-\pi^2}{4(\pi + 4)} + 2 \ln |\frac{\pi + 4}{4}|$

$I(\frac{\pi}{4}) = \frac{-\pi^2}{4(\pi + 4)} + \ln (\frac{(\pi + 4)^2}{16})$