Question

$\int \left(\frac{{\cos }^{3}x+{\cos }^{5}x}{{\sin }^{2}x+{\sin }^{4}x}\right)dx=$

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

Answer 1

The correct option is A $\sin x-\frac{2}{\sin x}-6{\tan }^{-1}\left(\sin x\right)+c$

$\int \frac{(co{s}^{3}x+co{s}^{5}x)}{(si{n}^{2}x+si{n}^{4}x)}dx$

$=\int \frac{(co{s}^2+co{s}^{4}x)}{(si{n}^{2}x+si{n}^{4}x)}d\left(sin\right)$

Substitute $\sin x=t$

$=\int \frac{(1-t^2)+(1+t^4-2t^2)}{t^2+t^4}dt$

$=\int \frac{t^4-3t^2+2}{t^2(t^2+1)}dt$

$\int [1+\frac{(2-4t^2)}{t^2(t^2+1)}]dt$

$\int 1.dt+2\int \frac{(t^2+1)-(t{)}^2}{t^2(t^2+1)}-dt4\int \frac{1}{t^2+1}dt$

$\int 1.dt+2\int \frac{1}{t^2}dt-2\int \frac{1}{t^2+1}dt-4\int \frac{1}{t^2+1}dt$

$=t+\frac{-2}{t}-6ta{n}^{-1}\left(t\right)+c$

$=sinx-2/sinx-6ta{n}^{-1}\left(sinx\right)+c$