Calculate trigonometric integral: $\int \sin(x)[\sec(2x)]^{3/2}dx$ I need some help to solve these integral:
$$\int \sin(x)[\sec(2x)]^{3/2}dx$$
Thank you.
 A: Firstly, note $$\sec 2x = \frac{1}{2\cos^2(x) - 1}$$ and hence 
$$\int \sin(x)[\sec(2x)]^{3/2} \ dx= \int \sin(x) \left[{2\cos^2(x) - 1}\right]^{-3/2} \ dx $$
Taking $u = \cos x$ yields 
$$-\int ({2u^2 - 1})^{-3/2} \ du$$
We can solve the above integral firstly for only nonnegative $u$ and use the fact that the function is even (hence its primitive odd) to extend it in the final step. Thus there is no problem with taking only the principal root when solving for $u$ in our next step; it all works out. 
Take $t = ({2u^2 - 1})^{-\frac 12}$ so that $$dt = \frac{-2u}{({2u^2 - 1})^{\frac 32}} \ du$$ and we get 
$$\int \frac{1}{2u} \ dt = \frac{1}{2\sqrt 2}\int \frac{{2t \ dt}}{\sqrt{1 + t^2}}  = \frac{1}{\sqrt{2}}\sqrt{1 + t^2} + C = \frac{u}{(2u^2 - 1)^{\frac 12}} + C = {\cos x}{\sqrt{\sec 2x}} + C$$
A: $$\int \sin(x)[\sec(2x)]^{3/2}dx=\int\frac{\sin x}{(\cos^2x-\sin^2x)^{3/2}}dx=\int\frac{\sin x}{(2\cos^2x-1)^{3/2}}dx$$
Now, let $u=\cos x$ and $du=-\sin x dx$
$$-\int \frac{du}{(2u^2-1)^{3/2}}$$
Now, let $u=\frac{\sec t}{\sqrt{2}}$ and $du=dt\frac{\sec t \tan t}{\sqrt{2}}$
$$-\frac{1}{\sqrt{2}}\int \frac{\sec t \tan t dt}{\tan^3 t}=-\frac{1}{\sqrt{2}}\int \frac{\sec t  dt}{\tan^2 t}=-\frac{1}{\sqrt{2}}\int \frac{\cos t  dt}{\sin^2 t}.$$
Finally, let $s=\sin t$ and $ds=\cos tdt$
Can you proceed from here? It seems that the back substitutions may be more difficult than the antiderivative.
