$\int \frac{a + b \sin x}{{(b + a \sin x)}^{2}} d x$

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$\frac{- \cos x}{b + a \sin x} + c$

$\frac{- \sin x}{b + a \sin x} + c$

$\frac{\sin x}{b + a \sin x} + c$

$\frac{\cos x}{b + a \sin x} + c$

answer is A.

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Detailed Solution

$\begin{array}{l} \int \frac{a + bsinx}{{(b + asinx)}^{2}} dx \\ \int \frac{a \sec^{2} x + b \tan x \sec x}{{(b \sec x + a tanx)}^{2}} \\ (multiply Nr . and Dr . by \sec^{2} x) \end{array}$
$\begin{array}{rcl} Put b \sec x + a \tan x & = & t \\ \Rightarrow (b \sec x \tan x + a \sec^{2} x) dx & = & dt \end{array}$
$\begin{array}{rcl} I & = & \int \frac{dt}{t^{2}} \\ = & \frac{- 1}{t} + c \\ = & \frac{- 1}{b \sec x + a \tan x} + c \\ = & \frac{- \cos x}{b + a \sin x} + c \end{array}$