If $\int \frac{\mathrm{d} x}{1+3 \sin ^2 x}=\frac{1}{2} an ^{-1}(\mathrm{f}(x))+\mathrm{c}$, where c is -a constant of integration, then $\mathrm{f}(x)$ is equal to

Question

If $\int \frac{\mathrm{d} x}{1+3 \sin ^2 x}=\frac{1}{2} 	an ^{-1}(\mathrm{f}(x))+\mathrm{c}$, where c is -a constant of integration, then $\mathrm{f}(x)$ is equal to

Aicharya · Accepted Answer

$\begin{aligned} & \int \frac{\mathrm{d} x}{1+3 \sin ^2 x}=\int \frac{\mathrm{d} x}{\sin ^2 x+\cos ^2 x+3 \sin ^2 x} \ & =\int \frac{\mathrm{d} x}{4 \sin ^2 x+\cos ^2 x} \ & =\int \frac{\sec ^2 x \mathrm{~d} x}{4 	an ^2 x+1} \ & =\frac{1}{4} \int \frac{\sec ^2 x \mathrm{~d} x}{	an ^2 x+\frac{1}{4}} \ & 	ext { Put } \mathrm{t}=	an x \Rightarrow \mathrm{dt}=\sec ^2 x \mathrm{~d} x \ & 	herefore \quad \int \frac{\mathrm{~d} x}{1+3 \sin ^2 x}=\frac{1}{4} \int \frac{\mathrm{dt}}{\mathrm{t}^2+\left(\frac{1}{2}ight)^2}=\frac{1}{4} \cdot 2 	an ^{-1}(2 \mathrm{t})+\mathrm{c} \ & =\frac{1}{2} 	an ^{-1}(2 \mathrm{t})+\mathrm{c} \ & =\frac{1}{2} 	an ^{-1}(2 	an x)+\mathrm{c} \ & 	herefore \quad \mathrm{f}(x)=2 	an x \end{aligned}$

If $\int \frac{\mathrm{d} x}{1+3 \sin ^2 x}=\frac{1}{2} \tan ^{-1}(\mathrm{f}(x))+\mathrm{c}$, where c is -a…

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