Question

$\int \frac{{\sin }^{3}xdx}{(1+{\cos }^{2}x)\sqrt{1+{\cos }^{2}x+{\cos }^{4}x}}dx$ is equal to

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

Answer 1

Answer: (A)

${\sec }^{-1}(\sec x+\cos x)+c$

$I=\int \frac{{\sin }^{3}xdx}{(1+{\cos }^{2}x)\sqrt{1+{\cos }^{2}x+{\cos }^{4}x}}dx$

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

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

Substitute $\sec x+\cos x=z\Rightarrow \frac{{\sin }^{3}x}{{\cos }^{2}x}dx=dz$

$\therefore I=\int \frac{dz}{z\sqrt{z^2-1}}={\sec }^{-1}z+c={\sec }^{-1}(\sec x+\cos x)+c$