Sorry, I'm new and I don't know how to show mathematical symbols on here. The integral from 0 to pi/2 of sin^4x dx. The exponent is just 4, not 4x. Thanks a lot!
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Originally Posted by element Sorry, I'm new and I don't know how to show mathematical symbols on here. The integral from 0 to pi/2 of sin^4x dx. The exponent is just 4, not 4x. Thanks a lot! $\displaystyle \int_0^{\frac{\pi}{2}} \sin^4 x\, dx = \int_0^{\frac{\pi}{2}} \sin^2 x(1 - \cos^2x)\, dx $ you can use the following identities: $\displaystyle \cos 2x = \cos^2 x - \sin^2x = 2\cos^2x - 1 = 1 - 2\sin^2x$
Originally Posted by kalagota $\displaystyle \int_0^{\frac{\pi}{2}} \sin^4 x\, dx = \int_0^{\frac{\pi}{2}} \sin^2 x(1 - \cos^2x)\, dx $ you can use the following identities: $\displaystyle \cos 2x = \cos^2 x - \sin^2x = 2\cos^2x - 1 = 1 - 2\sin^2x$ Uhm...why rewrite it as that? $\displaystyle \int\sin^4(x)dx=\int\left(\frac{1-\cos(2x)}{2}\right)^2dx$ Expand that and see that $\displaystyle \cos^2(nx)=\frac{1+\cos(2nx)}{2}$
Last edited by Mathstud28; Jun 23rd 2008 at 06:14 PM.
Originally Posted by Mathstud28 Uhm...why rewrite it as that? $\displaystyle \int\sin^4(x)dx=\int\left(\frac{1-\cos(2x)}{2}\right)dx$ Expand that and see that $\displaystyle \cos^2(nx)=\frac{1+\cos(2nx)}{2}$ How is $\displaystyle sin^4(x) = \frac{1 - cos(2x)}{2}$? -Dan
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