Question

The value of $\int \limits \frac {cos^3x+cos^5x}{sin^2x+sin^4x}dx \, is$

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

Answer 1

Answer: (C)

Let $I= \int \limits \frac {cos^3 x + cos^5 x}{sin^2x+sin^4 x}dx$
$ = \int \limits \frac {(cos^2x+cos^4x).cos \, x dx}{(sin^2x+sin^4x)}$
Put $sin \, x= t \Rightarrow cos \, x \, dx =dt $
$\therefore I=\int \limits \frac {[(1-t^2)+(1-t^2)^2]}{t^2+t^4}dt$
$\Rightarrow I=\int \limits \frac {1-t^2+1-2t^4+t^4}{t^2+t^4}dt$
$\Rightarrow I=\int \limits \frac {2-3t^2+t^4}{t^2(t^2+1)}dt ...(i)$
Using partial fraction for
$ \frac {y^2-3y+2}{y(y+1)}=1+ \frac {A}{Y}+ \frac {B}{y+1} [where,\, y=t^2]$
$\Rightarrow A=2, B=-6$
$\therefore \frac {y^2-3y+2}{y(y+1)}=1+ \frac {2}{y}- \frac {6}{y+1}$
Now, Eq. (i) reduces to, $I= \int \limits \bigg ( 1+ \frac {2}{t^2}- \frac {6}{1+t^2} \bigg )dt$
$ =t- \frac {2}{t}-6 tan^{-1}(t)+ c$
$ =sin \, x- \frac {2}{sin \, x}-6 tan^{-1}(sin \, x ) + c$