I'm usually pretty good with definite integrals, but this one's got me completely lost. Any help is appreciated! (And the sooner the better, please!)
$$\int_0^{\pi/2}\cos^{8}x\,dx.$$
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Yes use reduction formulae or trig identities... – bjd2385 Feb 26 '15 at 02:42
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possible duplicate of Prove $\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$ – Feb 26 '15 at 03:01
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This is a Wallis integral. – Lucian Feb 26 '15 at 05:34
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We have, $$\int_{0}^{\pi /2}\cos^8x\,dx$$
$$=\int_{0}^{\pi /2}\sin^0x \cos^8x\,dx$$
$$=\frac{\Gamma\left(\frac{0+1}{2}\right)\Gamma\left(\frac{8+1}{2}\right)}{2.\Gamma\left(\frac{0+8+2}{2}\right)}$$
$$=\frac{\Gamma(1/2)\Gamma(9/2)}{2.\Gamma(5)}$$
$$=\frac{35\pi}{2^8}.$$
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Hint: $\cos^8(x)$ can be written as :$$(\cos^2(x))^4=\left(\frac12\left(1+\cos(2x)\right)\right)^4=\frac{1}{16}\cos^4(2x)+\frac14\cos^3(2x)+\frac38\cos^2(2x)+\frac14\cos(2x)+\frac{1}{16}$$
Now, keep using the half-angle identity on the terms until no powers remain. Integrating should be straight-forward after that.
You will then have: $$\frac{1}{128}\int_{0}^{\pi/2}(56\cos(2x)+28\cos(4x)+8\cos(6x)+\cos(8x)+35)dx$$