The value of ∫cos x(1+sin x)(2+sin x)dx is

Let, I = ∫cos x(1+sinx)(2+sin x)dxPut 1+sin⁡x=t⇒cos⁡xdx=dt Now ∫cos x(1+sinx)(2+sin x)dx=[using partial fractions]=∫dtt−∫dt1+t=log⁡|t|−log⁡|1+t|+C=log⁡t1+t+C ∵log⁡m−log⁡n=log⁡mn =log⁡1+sin⁡x2+sin⁡x+C

The value of ∫cos x(1+sin x)(2+sin x)dx is

Let, I = ∫cos x(1+sinx)(2+sin x)dxPut 1+sin⁡x=t⇒cos⁡xdx=dt Now ∫cos x(1+sinx)(2+sin x)dx=[using partial fractions]=∫dtt−∫dt1+t=log⁡|t|−log⁡|1+t|+C=log⁡t1+t+C ∵log⁡m−log⁡n=log⁡mn =log⁡1+sin⁡x2+sin⁡x+C

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The value of $\int \frac{\cos x}{(1 + \sin x) (2 + \sin x)} dx$ is

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$\log |\frac{1 + \sin x}{2 - \sin x}| + C$

$\log |\frac{1 + \sin x}{2 + \sin x}| + C$

$\log |\frac{2 + \sin x}{1 - \sin x}| + C$

$\log |\frac{2 + \sin x}{1 + \sin x}| + C$

answer is B.

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Detailed Solution

Let, I = $\int \frac{\cos x}{(1 + sinx) (2 + \sin x)} dx$
Put $1 + \sin x = t \Rightarrow \cos xdx = dt$
$N o w \int \frac{\cos x}{(1 + sinx) (2 + \sin x)} dx =$
[using partial fractions]
$= \int \frac{dt}{t} - \int \frac{dt}{1 + t} = \log | t | - \log | 1 + t | + C$
$= \log |\frac{t}{1 + t}| + C [∵ \log m - \log n = \log \frac{m}{n}] = \log |\frac{1 + \sin x}{2 + \sin x}| + C$