Question
The value of the integral $$\int {\frac{{{{\cos }^3}x + {{\cos }^5}x}}{{{{\sin }^2}x + {{\sin }^4}x}}dx} $$ is-
A.
$$\sin \,x - 6\,{\tan ^{ - 1}}\left( {\sin \,x} \right) + c$$
B.
$$\sin \,x - 2{\left( {\sin \,x} \right)^{ - 1}} + c$$
C.
$$\sin \,x - 2{\left( {\sin \,x} \right)^{ - 1}} - 6\,{\tan ^{ - 1}}\left( {\sin \,x} \right) + c$$
D.
$$\sin \,x - 2{\left( {\sin \,x} \right)^{ - 1}} + 5\,{\tan ^{ - 1}}\left( {\sin \,x} \right) + c$$
Answer :
$$\sin \,x - 2{\left( {\sin \,x} \right)^{ - 1}} - 6\,{\tan ^{ - 1}}\left( {\sin \,x} \right) + c$$
Solution :
$$\eqalign{
& {\text{Let,}}\,\,\,\,\,I = \int {\frac{{{{\cos }^3}x + {{\cos }^5}x}}{{{{\sin }^2}x + {{\sin }^4}x}}dx} \cr
& = \int {\frac{{\left( {{{\cos }^2}x + {{\cos }^4}x} \right)\cos \,x}}{{{{\sin }^2}x\left( {1 + {{\sin }^2}x} \right)}}dx} \cr
& = \int {\frac{{\left[ {1 - {{\sin }^2}x + {{\left( {1 - {{\sin }^2}x} \right)}^2}} \right]\cos x}}{{{{\sin }^2}x\left( {1 + {{\sin }^2}x} \right)}}dx} \cr
& = \int {\frac{{\left( {2 - 3\,{{\sin }^2}x + {{\sin }^4}x} \right)\cos \,x}}{{{{\sin }^2}x\left( {1 + {{\sin }^2}x} \right)}}dx} \cr
& {\text{Put }}\sin \,x = t\,\,\, \Rightarrow \cos \,x\,dx = dt \cr
& I = \int {\frac{{2 - 3{t^2} + {t^4}}}{{{t^4} + {t^2}}}dt,\,\,\,} I = \int {\left( {1 + \frac{2}{{{t^2}}} - \frac{6}{{{t^2} + 1}}} \right)dt\,} \cr
& = t - \frac{2}{t} - 6\,{\tan ^{ - 1}}\left( t \right) + C \cr
& = \sin \,x - 2{\left( {\sin \,x} \right)^{ - 1}} - 6\,{\tan ^{ - 1}}\left( {\sin \,x} \right) + C \cr} $$