Question

Integrate $\frac{1+sinx}{sinx+sinxcosx}$

Answer 1

I = ( 1 + sinx)dx /( sinx + sinx .cosx)
we Know ,
sin2x = 2tanx/( 1+ tan²x)
cos2x = (1 - tan²x)/( 1 + tan²x)

use this

I = { 1 + 2tanx/2/( 1+ tan²x/2)}dx/{ 2tanx/2/( 1+tan²/2.) + 2tanx/2( 1 - tan²x/2)/( 1+ tan²x/2)² }

= ( 1 + tan²x/2 +2tanx/2)dx/( 2tanx/2)( 1+ tan²x/2 + 1 - tan²x/2)/( 1 + tan²x/2)

= ( 1 + tan²x/2)( 1 + tanx/2)²dx/2tanx/2× 2

= (1/4) { (sec²x/2)( 1 +tanx/2)dx/tanx/2

[ note :- ( 1 + tan²x/2 ) = sec²x/2 ]

now , let tanx/2 = P
differentiate
sec²x/2 × 1/2 dx = dP
sec²x/2dx = 2dP
use this in above

I = ( 1 + P)²2.dP/4P
=1/2 × ( 1+P² + 2P)dP/P
=1/2 × { dP/P + P.dP + 2dP }
= 1/2× { lnP + P²/2 + 2P }

put P = tanx/2

I = 1/2tanx/2 + 1/4tan²x/2 + tanx + C

Answer 2

hope it helps you......

pls mark as the brainliest answer...

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