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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)

12 years ago

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