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[i] ∫(x+sin x) / (1+ cos x)
[ii] ∫cos -1( ( 1-x2)/( 1+x2) ) [ why can't we put x=tan θ and dx=sec2 θ ->i'm not getting the answer by this way ]
[iii] ∫sin -1√(x /a+x)
hi,
i) ∫(x+sin x) / (1+ cos x) =
= ∫sin x / (1+ cos x)dx + ∫x / (1+ cos x)dx
= - ln ( 1+ cos x ) + ∫x / (1+ 2 cos^2 x/2 -1) dx
= - ln ( 1+ cos x ) + 1/2 ∫x sec^2 x/2 dx
= - ln ( 1+ cos x ) + 1/2 [ 2x tan x/2 - 2∫ tan x/2 dx ]
= - ln ( 1+ cos x ) + 1/2 [ 2x tan x/2 + 4 ln cos x/2]
= - ln ( 1+ cos x ) + x tan x/2 + 2 ln cos x/2
ii) ∫cos -1( ( 1-x2)/( 1+x2) ) put x=tan t and dx= sec2 t dt
= ∫cos -1( ( 1 - tan^2 t ) /( 1 + tan ^2 t ) dt
= ∫cos -1( cos ^2 t - sin^2 t ) / ( cos ^2 t + sin^2 t ) sec2 t dt
= ∫cos -1( cos 2t ) sec2t dt
= ∫ 2t sec2 t dt
= 2 [ t tant - ∫ tant dt ]
= 2 [ t tant + ln cost ]
= 2 [ x. tan^-1 x + ln cos ( tan^-1 x ) ]
= 2 [ x. tan^-1 x + ln cos { cos^-1 1/ √1+ x^2 }
= 2 [ x. tan^-1 x + ln ( 1/ √1+ x^2) ]
iii) ∫sin -1√(x /a+x)
put x = ay , dx = a dy
= a ∫sin -1√ y/1+y dy , put y = tan ^2 t , dy = 2 tant sec^2 t dt
= a ∫sin -1√ tan ^2 t / 1+ tan ^2 t ) 2 tant sec^2 t dt
= 2 a ∫sin -1√ sin^2 t . sin t / cos^3 t dt
= 2 a ∫ t . sin t / cos^3 t dt
= 2a [ 1/2 t . 1/ cos^2 t - 1/2 ∫ 1/ cos^2 t dt ]
= 2a [ 1/2 t . 1/ cos^2 t - 1/2 tan t ]
= a t . 1/ cos^2 t - a tan t
= a tan^-1 √ x/a . 1/ cos^2 ( tan^-1 √ x/a ) - a √ x/a
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