Question

please help me with this qn guys
​

Answer 1

Step-by-step explanation:

We have,

$\int^{ \frac{\pi}{2} } _{0} \frac{1}{1 + \sin(x) } dx$

$= \int^{ \frac{\pi}{2} } _{0} \frac{1}{1 + \frac{2 \tan( \frac{x}{2} ) }{1 + \tan^{2} ( \frac{x}{2} ) } } dx$

$= \int^{ \frac{\pi}{2} } _{0} \frac{1 + \tan^{2} ( \frac{x}{2} ) }{1 + \tan^{2} ( \frac{x}{2} ) + 2 \tan( \frac{x}{2} ) } dx$

$= \int^{ \frac{\pi}{2} } _{0} \frac{ \sec^{2} ( \frac{x}{2} ) }{(1 + \tan ( \frac{x}{2} ) )^{2} } dx$

Now, put

$1 + \tan( \frac{x}{2} ) = y \\ \implies \: \sec^{2} ( \frac{x}{2} ) dx = 2dy$

And limits will be,

$y = 1 + \tan( \frac{\pi}{4} ) = 2 \\ y = 1 + \tan( \frac{0}{2} ) = 1$

so,

$= \int^{ 2 } _{1} \frac{ 2 }{y ^{2} } dy$

$= - 2 [ \frac{1}{y} ]^{2} _{1}$

$= - 2 ( \frac{1}{2} - 1)$

$= ( - 2) \times \frac{1}{( - 2)}$

$= 1$