JEE Main 2012 — Mathematics Calculus

$$\begin{aligned} & \text { Let } f(x)=\int\left(\frac{x^2+\sin ^2 x}{1+x^2}\right) \sec ^2 x d x \\ & =\int \frac{x^2 \sec ^2 x+\frac{\sin ^2 x}{\cos ^2 x}}{1+x^2} d x \end{aligned}$$​ Let f(x)=∫(1+x2x2+sin2x​)sec2xdx=∫1+x2x2sec2x+cos2xsin2x​​dx​ $$\begin{aligned} & =\int \frac{x^2 \sec ^2 x+\tan ^2 x}{1+x^2} d x \\ & =\int \frac{x^2\left(1+\tan ^2 x\right)+\tan ^2 x}{1+x^2} d x \\ & =\int \frac{x^2+\tan ^2 x\left(1+x^2\right)}{1+x^2} d x \\ & =\int \frac{x^2}{1+x^2} d x+\int \tan ^2 x d x \\ & =\int \frac{x^2+1-1}{1+x^2} \mathrm{dx}+\int\left(\sec ^2 \mathrm{x}-1 \mathrm{dx}\right. \\ & =\int 1 d x-\int \frac{d x}{1+x^2}+\int \sec ^2 x d x-\int d x \\ & =-\tan -1 x+\tan ^{-1} x+\mathrm{c} \\ & \text { Given: } f(0)=0 \\ & \Rightarrow f(0)=-\tan ^{-1} 0+\tan ^2+c \\ & \Rightarrow c=0 \\ & \therefore f(x)=-\tan ^{-1} x+\tan ^2 x \end{aligned}$$​=∫1+x2x2sec2x+tan2x​dx=∫1+x2x2(1+tan2x)+tan2x​dx=∫1+x2x2+tan2x(1+x2)​dx=∫1+x2x2​dx+∫tan2xdx=∫1+x2x2+1−1​dx+∫(sec2x−1dx=∫1dx−∫1+x2dx​+∫sec2xdx−∫dx=−tan−1x+tan−1x+c Given: f(0)=0⇒f(0)=−tan−10+tan2+c⇒c=0∴f(x)=−tan−1x+tan2x​ Now, $$f(1)=-\tan ^{-1}(1)+\tan 1=\tan 1-\frac{\pi}{4}$$f(1)=−tan−1(1)+tan1=tan1−4π​