Mastering Trig Subs: Simplify \int\cos^5(x)dx without the Headache

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These trig subs are killing me.

\int\cos^5(x)dxhints?
 
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\cos^{5} x = \left(1 - \sin^2 x \right)^{2} \cos x

Regards,
George
 
Thanks George... I am slowly getting the hang of this. I appreciate your help.
 
One of the first things you should have learned: if you have sine or cosine to an odd power, factor out one of them to use with the dx.

cos5 x dx= (cos4 x)(cos x dx)
= (cos2 x)2 (cos x dx)= (1- sin2 x)2(cos x dx)

Now, let u= sin x.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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