$\int \frac{cos x}{{sin}^{2} x + 4 sin x + 5} d x = {tan}^{- 1} [sin x + k] .$ Find the value of $k$ .

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Solution

Let $I = \int \frac{cos x}{{sin}^{2} x + 4 sin x + 5} d x$
Put $sin x = t \Rightarrow cos x d x = d t$
We get
$I = \int \frac{d t}{t^{2} + 4 t + 5} = \int \frac{d t}{{(t + 2)}^{2} + 1}$
$= {tan}^{- 1} \frac{(t + 2)}{1} = {tan}^{- 1} [sin x + 2]$
Thus the value of $k = 2$ .

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