# $\int \frac{dx}{1+\sin x}$ Option 1) $\frac{1}{1+\tan _{\frac{x}{2}} }+C$ Option 2) $-\frac{1}{1+\tan _{\frac{x}{2}} }+C$ Option 3) $\frac{1}{1+\cot _{\frac{x}{2}} }+C$ Option 4) $-\frac{1}{1+\cot _{\frac{x}{2}} }+C$

G gaurav

As we learned,

Type of Integration by perfect square -

 The integrals are of the from  (i)    $\int \frac{1}{a\cos x+b\sin x}dx$ (ii) $\int \frac{1}{a+b\cos x}dx$ (iii) $\int \frac{1}{a+b\sin x}dx$

- wherein

Working rule :

Resolve :

$\cos x=\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2}$

and

$\sin x=2\sin\frac{x}{2}\cos\frac{x}{2}$

$\sin x=2\sin\frac{x}{2}\cos\frac{x}{2}$

or $\sin x=\frac{2\tan _{\frac{x}{2}}}{1+\tan _{\frac{x}{2}}^{2}}$

Thus, $I=\int \frac{\sec _{\frac{x}{2}}^{2}dx}{1+\tan _{\frac{x}{2}}^{2}+2\tan _{\frac{x}{2}}}=\int \frac{dt}{\left ( 1+t \right )^{2}}=\frac{-1}{1+\tan _{\frac{x}{2}}}+C$

Option 1)

$\frac{1}{1+\tan _{\frac{x}{2}} }+C$

Option 2)

$-\frac{1}{1+\tan _{\frac{x}{2}} }+C$

Option 3)

$\frac{1}{1+\cot _{\frac{x}{2}} }+C$

Option 4)

$-\frac{1}{1+\cot _{\frac{x}{2}} }+C$

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