Question

The value of $\int \frac{co{s}^{3}x+co{s}^{5}x}{si{n}^{2}x+si{n}^{4}x}dxis$

• A. $sinx-6ta{n}^{-1}(sinx)+c$
  
• B. $sinx-2(sinx{)}^{-1}+c$
  
• C. $sinx-2(sinx{)}^{-1}-6ta{n}^{-1}(sinx)+c$
  
• D. $sinx-2(sinx{)}^{-1}+5ta{n}^{-1}(sinx)+c$
  

Answer 1

Answer: (C)

Let $I=\int \frac{co{s}^{3}x+co{s}^{5}x}{si{n}^{2}x+si{n}^{4}x}dx$
$=\int \frac{(co{s}^{2}x+co{s}^{4}x).cosxdx}{(si{n}^{2}x+si{n}^{4}x)}$
Put $sinx=t\Rightarrow cosxdx=dt$
$\therefore I=\int \frac{[(1-t^2)+(1-t^2{)}^2]}{t^2+t^4}dt$
$\Rightarrow I=\int \frac{1-t^2+1-2t^4+t^4}{t^2+t^4}dt$
$\Rightarrow I=\int \frac{2-3t^2+t^4}{t^2(t^2+1)}dt...(i)$
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, $I=\int (1+\frac{2}{t^2}-\frac{6}{1+t^2})dt$
$=t-\frac{2}{t}-6ta{n}^{-1}(t)+c$
$=sinx-\frac{2}{sinx}-6ta{n}^{-1}(sinx)+c$