, ^3 ,x , , + , , ^5 ,x ^2 ,x , , + , , ^4 ,x ,dx - Vidyadip

MCQ

$\int {\,\frac{{{{\cos }^3}\,x\,\, + \,\,{{\cos }^5}\,x}}{{{{\sin }^2}\,x\,\, + \,\,{{\sin }^4}\,x}}} \,dx$

A
$sin\, x - 6\, \tan^{-1} (\sin \,x) + c$
B
$\sin\, x - 2\, \sin^{-1} x + c$
✓
$\sin\, x - 2 (\sin\, x)^{-1} - 6 \tan^{-1} (\sin x) + c$
D
$\sin\, x - 2 \,(\sin \,x)^{-1} + 5 \tan^{-1}\, (\sin x) + c$

✓

Answer

Correct option: C.

$\sin\, x - 2 (\sin\, x)^{-1} - 6 \tan^{-1} (\sin x) + c$

c
$\sin\, x = t$ ;

$I = \int {\,\frac{{\left( {1\, - \,{t^2}} \right)\,\,\left( {2\, - \,{t^2}} \right)}}{{{t^2}\,\,\left( {1\, + \,{t^2}} \right)}}} $

$dt =\int {\,\frac{{(y - 1)\,\,(y\, - \,2)}}{{y\,(1\, + \,y)}}} dy$

$ = 1 +\frac{{2\,(1\,\, - \,\,2y)}}{{y\,(y\, + \,1)}}$ ; $(y = t^2)$
$= 1 + 6 \left[ {\frac{1}{{3\,y}}\,\, - \,\,\frac{1}{{y\, + \,1}}} \right]$

$= \left( {1\,\, + \,\,\frac{2}{{{t^2}}}\,\, - \,\,\frac{6}{{1\, + \,{t^2}}}} \right) \,dt$

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