Question

$\int \frac{\cos x}{{\sin }^{2}x+4\sin x+5}dx$ is equal to

• A. ${\sin }^{-1}(\sin x+2)+c$
• B. ${\tan }^{-1}(\sin x+2)+c$
• C. ${\cot }^{-1}(\sin x+2)+c$
• D. ${\sec }^{-1}(sinx+2)+c$

Answer 1

The correct option is B ${\tan }^{-1}(\sin x+2)+c$

$\int \frac{d\left(\sin x\right)}{{\sin }^2+4\sin x+4+1}$

$=\int \frac{d\left(\sin x\right)}{(\sin x+2{)}^2+1}$

$={\tan }^{-1}(\sin x+2)+c$ ....... $\int \frac{1}{a^2+x^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+c$