A clarification on a step in an integration question

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Homework Statement



I was given this question as a part of an assignment and lost a mark because of a step.

Homework Equations


the integral of
cos^5(x) dx

after some fiddling and substitution it gets to this

(1 - u^2)^2 du
In the solutions there is a step that says
refine
= (u^2 - 1)^2
basically switching the the 1 and the u^2 around.

The Attempt at a Solution


Is this possible and if so how?
 
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It is possible because ##(-1)^2 = 1## and the distributive property of multiplication:$$ \eqalign { (a^2-1)^2 & = 1 * (a^2-1)^2 \\ & = (-1)^2 * (a^2-1)^2 \\ & = ( -1*(a^2-1) ) * ( -1*(a^2-1) ) \\ & = ( -1*(a^2-1) )^2 \\ & = ( 1 - a^2 )^2 } $$

Or, simply, because ## a^2 = (-a)^2 ##...:rolleyes:
 
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Good point. I'll have to remember that.
 
mattyk said:

Homework Statement



I was given this question as a part of an assignment and lost a mark because of a step.

Homework Equations


the integral of
cos^5(x) dx

after some fiddling and substitution it gets to this

(1 - u^2)^2 du
In the solutions there is a step that says
refine
= (u^2 - 1)^2
basically switching the the 1 and the u^2 around.

The Attempt at a Solution


Is this possible and if so how?

I don't know why you lost a mark. Both (1-u^2)^2 and (u^2 - 1)^2 are equal to u^4 - 2 u^2 + 1, so IF you performed that expansion you should not have lost a mark.
 
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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