Trouble with Integrating \cos^{5}7x\sin7x Using Substitution

AStaunton
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trying to solve the following integral by substitution but having trouble:

\int\cos^{5}7x\sin7xdx

I attempted to set u=\cos^{5}7x and ended up with (by chain rule...which I hope is correct!):

du=-35\cos^{4}7x\sin7xdx

This doesn't seem too helpful but can't think of a better substitution.

Any advice is appreciated

Andrew
 
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Try using two substitutions. Problems are often made to work out in the end, so try to think of what substitution(s) you can make so things will start to simplify nicely. What if you first set u = 7x first? Beautiful things might start to happen. ;)
 
Well, there's need for only one substitution. \cos 7 x =u
 
True that, I tend to break it down more. Though I suppose you can make the argument that the more you break it down the more you have to put together.
 
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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