Question

If $I=\int \frac{dx}{(1+\sin x{)}^4}=\frac{-A}{4992}(\frac{1}{7}{u}^7+\frac{3}{5}{u}^5+u^3+u)+C$, where $u=\frac{1-2\sin x}{1+\sin x}$ then $A$ is equal to

Answer 1

Let $I=\int \frac{1}{{(\sin x+1)}^4}dx$

Substituting $t=\tan \frac{x}{2}$

$\Rightarrow dt=\frac{1}{2}{\sec }^2\frac{x}{2}dx$

$\sin x=\frac{2t}{t^2+1},\cos x=\frac{1-t^2}{t^2+1},dx=\frac{2dt}{t^2+1}$

$I=\int \frac{2}{(t^2+1){(\frac{2t}{t^2+1}+1)}^4}dt=2\int \frac{{(t^2+1)}^3}{{(t+1)}^8}dt$

$=8\int ({\left(\frac{{(t-2)}^2}{t^2+2}\right)}^6+3{\left(\frac{{(t-2)}^2}{t^2+2}\right)}^4+3{\left(\frac{{(t-2)}^2}{t^2+2}\right)}^2+1)\left(\frac{-(1-t^2)(t^2+1)}{{(2t+1)}^2}\right)dt$

Let $\frac{{(t-2)}^2}{t^2+2}=u\Rightarrow \frac{1-2\sin x}{\sin x+1}=u$

$I=-8\int (u^6+3u^4+3u^2+1)du=-8(\frac{u^7}{7}+\frac{3}{5}{u}^4+u^3+u)$

Comparing with, $I=\int \frac{dx}{(1+\sin x{)}^4}=\frac{-A}{4992}(\frac{1}{7}{u}^7+\frac{3}{5}{u}^5+u^3+u)+C$

Therefore, $A=624$