Need help with integral and change the variable I have
$$\displaystyle \int_0^\pi \sqrt{x} ~\cos x ~dx$$
and I need to make change of the variable $u = \sin x$.
 A: $\int_0^\pi\sqrt{x}\cos x~dx=\int_0^\pi x^{\frac{1}{2}}\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n}}{(2n)!}dx=\int_0^\pi\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+\frac{1}{2}}}{(2n)!}dx=\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+\frac{3}{2}}}{(2n)!\left(2n+\dfrac{3}{2}\right)}\right]_0^\pi=\sum\limits_{n=0}^\infty\dfrac{2(-1)^n\pi^{2n+\frac{3}{2}}}{(2n)!(4n+3)}$
A: Hint: Try this if you want:
$$x\sqrt { x } =u\\ x={ u }^{ \frac { 2 }{ 3 }  }\\ \sqrt { x } ={ u }^{ \frac { 1 }{ 3 }  }\\ dx=\frac { 2 }{ 3 } { u }^{ \frac { -1 }{ 3 }  }du\\ \int _{ 0 }^{ \pi  } \sqrt { x } \cos  x~ dx=\int _{ 0 }^{ \pi  }{ u } ^{ \frac { 1 }{ 3 }  }\cos { u } ^{ \frac { 2 }{ 3 }  }\frac { 2 }{ 3 } { u }^{ \frac { -1 }{ 3 }  }du=\frac { 2 }{ 3 } \int _{ 0 }^{ \pi\sqrt\pi  }{ \cos { u } ^{ \frac { 2 }{ 3 }  }du } $$
$\int { \cos { u } ^{ \frac { 2 }{ 3 }  }du } $ does not have an analytical expression. You can try on wolframalpha.com .
A: Write x as a function of u: you have $u=\sin(x)$ so $x=\arcsin(u)$. It gets a little sticky when you have to simplify $\cos(\arcsin(u))$; remember the identity $\sin^2\theta+\cos^2\theta=1$, letting $\theta=\arcsin u$ (arcsin gives you an angle, recall) you get $\cos(\arcsin(u))=\sqrt{1-u^2}$.
Actually, this integral doesn't have a closed form; are you sure the problem wasn't $\int \sqrt{x} \cos^2 x dx$ 
