Question

$\int_{}^{}{\frac{\cos x}{\sin^{2}x + 4\sin x + 5}dx}$equals

• A. $\tan^{- 1}(\sin x) + c$
• B. $\tan^{- 1}(\sin x + 2) + c$
• C. $\tan^{- 1}(\sin x + 1) + c$
• D. None of these

Answer 1

Answer: (B)

$\tan^{- 1}(\sin x + 2) + c$

Answer 2

I =$\int_{}^{}{\frac{\cos x}{\sin^{2}x + 4\sin x + 5}dx} = \int_{}^{}\frac{\cos x}{(\sin x + 2)^{2} + 1}dx$

Put $\sin x + 2 = t \Rightarrow \cos xdx = dt$

I = $\int_{}^{}\frac{dt}{t^{2} + 1}$ $= \tan^{- 1}t + c = \tan^{- 1}(\sin x + 2) + c$.