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I have

$$\displaystyle \int_0^\pi \sqrt{x} ~\cos x ~dx$$

and I need to make change of the variable $u = \sin x$.

doraemonpaul
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user79727
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  • Why in the world make that change of variables? Let $u=\sqrt x$ instead! – Ted Shifrin May 27 '13 at 17:27
  • The teacher want this, he told me first to make change on the limits of the integral -pi/2 0 pi/2 :( – user79727 May 27 '13 at 17:33
  • Well, that would mean $u=\pi/2-x$, but this won't get you anywhere. Perhaps your teacher wants you to think about symmetry, like $\displaystyle\int_{-\pi}^{\pi} x\cos x,dx$, but that won't help here. – Ted Shifrin May 27 '13 at 17:37
  • This will not work. You'll end up with two integrals that can be done only by coming back to the original problem and doing it the right way!! Sorry to be a grouch. Sometimes — and I should know — teachers just make mistakes. – Ted Shifrin May 27 '13 at 17:52
  • Actually, your integrand doesn't have an elementary antiderivative; are you sure the problem wasn't $\int \sqrt{x} \cos^2 x dx$? – Zen May 27 '13 at 17:53
  • Yeah, @Zen wins. We get the Fresnel integral, and, although the improper integral $\int_0^\infty \sin(t^2),dt$ can be evaluated, this can't. As I said, we teachers make mistakes. – Ted Shifrin May 27 '13 at 18:01
  • This is not an elementary integral. – Mhenni Benghorbal May 28 '13 at 00:11

3 Answers3

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$\int_0^\pi\sqrt{x}\cos x~dx=\int_0^\pi x^{\frac{1}{2}}\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n}}{(2n)!}dx=\int_0^\pi\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+\frac{1}{2}}}{(2n)!}dx=\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+\frac{3}{2}}}{(2n)!\left(2n+\dfrac{3}{2}\right)}\right]_0^\pi=\sum\limits_{n=0}^\infty\dfrac{2(-1)^n\pi^{2n+\frac{3}{2}}}{(2n)!(4n+3)}$

doraemonpaul
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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 .

newzad
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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 $\int \sqrt{x} \cos^2 x dx$

Zen
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