Question

If $\int \frac{co{s}^{7}x+co{s}^{5}x}{si{n}^{2}x+si{n}^{4}x}dx=-\frac{A}{744}si{n}^{3}x+5sinx-\frac{2}{sinx}-12ta{n}^{-1}\left(sinx\right)+C$ then A is equal to

Answer 1

$I=\int \frac{{\cos }^{7}x+{\cos }^{5}x}{{\sin }^{2}x+{\sin }^{4}x}dx=-\int \frac{({\sin }^{2}x-2){({\sin }^{2}x-1)}^2\cos x}{{\sin }^{2}x+{\sin }^{4}x}dx$
Put $t=\sin x\Rightarrow dt=\cos xdx$
$I=-\int \frac{(t^2-2){(t^2-1)}^2}{t^4+t^2}dt=-\int (t^2+\frac{12}{t^2+1}-\frac{2}{t^2}-5)dt=-\frac{t^3}{3}+5t-\frac{2}{t}-12{\tan }^{-1}t+c=-\frac{{\sin }^{3}x}{3}+5\sin x-\frac{2}{\sin x}-12{\tan }^{-1}\left(\sin x\right)+c$
Therefore $A=248$