Q37 / 90 · JEE Main 2012 (19 May Online) · Mathematics
If $f(x)=\int\left(\frac{x^2+\sin ^2 x}{1+x^2}\right) \sec ^2 x d x$ and $f(0)=0$, then $f(1)$ equals
- Correct answerA$\tan 1-\frac{\pi}{4}$
- B$\tan 1+1$
- C$\frac{\pi}{4}$
- D$1-\frac{\pi}{4}$
Solution
$$
\begin{aligned}
& \text { Let } f(x)=\int\left(\frac{x^2+\sin ^2 x}{1+x^2}\right) \sec ^2 x d x \\
& =\int \frac{x^2 \sec ^2 x+\frac{\sin ^2 x}{\cos ^2 x}}{1+x^2} d x
\end{aligned}
$$
$$
\begin{aligned}
& =\int \frac{x^2 \sec ^2 x+\tan ^2 x}{1+x^2} d x \\
& =\int \frac{x^2\left(1+\tan ^2 x\right)+\tan ^2 x}{1+x^2} d x \\
& =\int \frac{x^2+\tan ^2 x\left(1+x^2\right)}{1+x^2} d x \\
& =\int \frac{x^2}{1+x^2} d x+\int \tan ^2 x d x \\
& =\int \frac{x^2+1-1}{1+x^2} \mathrm{dx}+\int\left(\sec ^2 \mathrm{x}-1 \mathrm{dx}\right. \\
& =\int 1 d x-\int \frac{d x}{1+x^2}+\int \sec ^2 x d x-\int d x \\
& =-\tan -1 x+\tan ^{-1} x+\mathrm{c} \\
& \text { Given: } f(0)=0 \\
& \Rightarrow f(0)=-\tan ^{-1} 0+\tan ^2+c \\
& \Rightarrow c=0 \\
& \therefore f(x)=-\tan ^{-1} x+\tan ^2 x
\end{aligned}
$$
Now,
$$
f(1)=-\tan ^{-1}(1)+\tan 1=\tan 1-\frac{\pi}{4}
$$
Hence, the correct answer is option: 'A'.