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

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Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { Let } f(x)=\int\left(\frac{x^2+\sin ^2 x}{1+x^2}ight) \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+	an ^2 x}{1+x^2} d x \ & =\int \frac{x^2\left(1+	an ^2 xight)+	an ^2 x}{1+x^2} d x \ & =\int \frac{x^2+	an ^2 x\left(1+x^2ight)}{1+x^2} d x \ & =\int \frac{x^2}{1+x^2} d x+\int 	an ^2 x d x \ & =\int \frac{x^2+1-1}{1+x^2} \mathrm{dx}+\int\left(\sec ^2 \mathrm{x}-1 \mathrm{dx}ight. \ & =\int 1 d x-\int \frac{d x}{1+x^2}+\int \sec ^2 x d x-\int d x \ & =-	an -1 x+	an ^{-1} x+\mathrm{c} \ & 	ext { Given: } f(0)=0 \ & \Rightarrow f(0)=-	an ^{-1} 0+	an ^2+c \ & \Rightarrow c=0 \ & 	herefore f(x)=-	an ^{-1} x+	an ^2 x \end{aligned} $ Now, $ f(1)=-	an ^{-1}(1)+	an 1=	an 1-\frac{\pi}{4} $

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

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