Question

lf $f\left(x\right)=\int \frac{(x^2+{\sin }^{2}x)}{1+x^2}{\sec }^{2}xdx$ and $f\left(0\right)=0$ then $f\left(1\right)$ is equal to

• A. $1-\frac{\pi }{4}$
• B. $\frac{\pi }{4}-1$
• C. $\tan 1-\frac{\pi }{4}$
• D. $\frac{\pi }{4}-\tan 1$

Answer 1

Answer: (C)

$\tan 1-\frac{\pi }{4}$

Given, $f\left(x\right)=\int \frac{(x^2+{\sin }^{2}x)}{1+x^2}{\sec }^{2}xdx$
$\Rightarrow$ $\int \frac{(x^2{\sec }^{2}x+{\tan }^{2}x)}{1+x^2}dx$
$\Rightarrow$ $\int \frac{\left(x^2\right(1+{\tan }^{2}x)+{\tan }^{2}x)}{1+x^2}dx$

$\Rightarrow$ $\int \frac{(x^2+1){\tan }^{2}x+x^2}{1+x^2}dx$
$\Rightarrow$ $\int {\tan }^{2}x+\frac{1+x^2}{1+x^2}-\frac{1}{1+x^2}dx$
$\Rightarrow$ $\tan x-{\tan }^{-1}x+c$ ,where $c$ is a constant.
Since, $f\left(0\right)=0$ we put $x=0$ in the above equation to get $c=0$ .
Hence , $f\left(1\right)=\tan 1-{\tan }^{-1}1=\tan 1-\frac{\pi }{4}$ is the answer.