If $\int \frac{\sin ^3 x\left( an ^{-1}(\sec x+\cos x) ight)^{-1}}{\left(\cos ^4 x+3 \cos ^2 x+1 ight)} d x=f(x)+C$, then $e^{\mathrm{f}(x)}=$

Question

If $\int \frac{\sin ^3 x\left(	an ^{-1}(\sec x+\cos x)ight)^{-1}}{\left(\cos ^4 x+3 \cos ^2 x+1ight)} d x=f(x)+C$, then $e^{\mathrm{f}(x)}=$

Aicharya · Accepted Answer

$ I=\int \frac{\sin ^3(x)}{\left(\cos ^4 x+3 \cos ^2 x+1ight)\left(	an ^{-1} \operatorname{cosec} x+\cos xight)} d x $ Let $	an ^{-1}(\sec x+\cos x)=t$ $ \begin{aligned} & \Rightarrow \frac{1}{1+(\sec x+\cos x)^2}[\sec x 	an x-\sin x] d x=d t \ & \Rightarrow \frac{\cos ^2 x}{\cos ^2 x+\cos ^4 x+2 \cos ^2 x+1} \ & {\left[\frac{\sin x-\sin x \cdot \cos ^2 x}{\cos ^2 x}ight] d x=d t} \ & =\frac{\sin x\left(1-\cos ^2 xight)}{\cos ^2 x+\cos ^4 x+2 \cos ^2 x+1} d x=d x \ & =\frac{\sin ^3 x}{\cos ^4 x+3 \cos ^2 x+1} d x=d t \ & =\int \frac{d t}{t} \ & =\log (t)+c \ & =\log \left|	an ^{-1} \operatorname{csec}^2(x)+\cos (x)ight|+c \ & \Rightarrow e^{f(x)}=	an ^{-1}(\sec x+\cos x)+c \end{aligned} $

If $\int \frac{\sin ^3 x\left(\tan ^{-1}(\sec x+\cos x)\right)^{-1}}{\left(\cos ^4 x+3 \cos ^2 x+1\right)} d…

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