Question

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

• A. ${\tan }^{-1}(\sec x+\cos x)+C$
• B. ${\log }_e|{\tan }^{-1}(\sec x+\cos x)|+C$
• C. $\frac{1}{(\sec x+\cos x{)}^2}+C$
• D. None of the above

Answer 1

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

Let $I=\int \frac{{\sin }^{3}x}{({\cos }^{4}x+3{\cos }^{2}x+1){\tan }^{-1}(\sec x+\cos x)}dx$
$=\int \frac{{\sin }^{3}x/{\cos }^{2}x}{({\cos }^{2}x+3+{\sec }^{2}x){\tan }^{-1}(\sec x+\cos x)}dx$
$=\int \left[\frac{1}{1+(\sec x+\cos x{)}^2}\times \frac{\sin x(1-{\cos }^{2}x)}{{\cos }^{2}x}\times \frac{1}{{\tan }^{-1}(\sec x+\cos x)}\right]dx$
$=\int \left[\frac{1}{{\tan }^{-1}(\sec x+\cos x)}\times \frac{1}{1+(\sec x+\cos x{)}^2}\times (\tan x\sec x-\sin x)\right]dx$
$=\int \frac{1}{{\tan }^{-1}(\sec x+\cos x)}\left\{{\tan }^{-1}\right(\sec x+\cos x\left)\right\}$
Therefore, $I={\log }_e|{\tan }^{-1}(\sec x+\cos x)+C$