# Change the limit of an integral

How can I change the limit of $\displaystyle\int\limits_0^\pi f(x)\,dx$ to $\displaystyle\int\limits_0^{\pi/2} f(x)\,dx$ ?

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If you mean "how can I compute $\int^{\pi/2}_0f(x)dx$ from $\int^\pi_0f(x)dx$?", then you can't without some conditions on $f$. The behaviour of $f$ in the two halves of the interval could be completely unrelated. –  Matthew Pressland Oct 16 '12 at 13:55
@Matt Pressland ...no...i just want the way to change the limit( 0 to $\pi$ )to( 0 to $\pi/2$) –  curious Oct 16 '12 at 13:57
We have that $\int_0^\pi f(x)dx = 2\int_0^{\pi/2} f(2x)dx$, but without any other information on $f$ that's the best you can do. –  Arthur Oct 16 '12 at 14:01
@Arthur....My problem is to solve $\int_o^\pi Sin^4{x}\,dx$ –  curious Oct 16 '12 at 14:03
Since $\sin^4(x)$ is symmetric about $x=\pi/2$ we have $\int_0^\pi \sin^4(x)dx = 2\int_0^{\pi/2} \sin^4(x)$. –  Arthur Oct 16 '12 at 14:08

$\int_o^\pi f(x)dx = \int_0^{\pi/2} 2 f(2y)dy$
by the change of variables $y := x/2$. Is this what you are looking for?