Integrate
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Hello mate, answer is here!!
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Step-by-step explanation:
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.
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