Question

$\int \frac{{\sin }^{3}x}{({\cos }^{4}x+3{\cos }^{2}x+1){\tan }^{-1}(\sec x+\cos x)}dx=$

• A. ${\tan }^{-1}(\sec x+\cos x)+c$
• B. $\log |{\tan }^{-1}(\sec x+\cos x)|+c$
• C. $\frac{1}{(\sec x+{\cos }^{2}x{)}^2}+c$
• D. $\log |\sec x+\cos x|+c$

Answer 1

The correct option is B $\log |{\tan }^{-1}(\sec x+\cos x)|+c$

Put ${\tan }^{-1}(\sec x+\cos x)=f\left(x\right)\Rightarrow f^{\prime }\left(x\right)=\frac{{\sin }^{3}x}{{\cos }^{4}x+3{\cos }^{2}x+1}\therefore \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\log |f\left(x\right)|+c\therefore \text{Integral}=\log |{\tan }^{-1}(\sec x+\cos x)|+c$