Is Using Exponent Identities Allowed in Solving Trigonometric Integrals?

XJellieBX
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Homework Statement


Compute \int^{\pi/2}_{0} \frac{sin^{2009}x}{sin^{2009}x + cos^{2009}x}

I used the identity cos^{2}= 1 - sin^{2}, but instead I set the exponent as 2009. And so I ended up with the answer being -1. I'm just wondering whether this is a legal solution or am I not allowed to do that. Thanks.
 
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sin^{2}x + cos^{2}x = 1 is the so called Pythagorean trigonometric identity. It is not valid when replacing the exponent 2 by another number, i.e.,

sin^{n}x + cos^{n}x \neq 1 for n\neq 2.
 
Thank you, I really needed that second opinion =)
 
XJellieBX said:

Homework Statement


Compute \int^{\pi/2}_{0} \frac{sin^{2009}x}{sin^{2009}x + cos^{2009}x}

I used the identity cos^{2}= 1 - sin^{2}, but instead I set the exponent as 2009. And so I ended up with the answer being -1. I'm just wondering whether this is a legal solution or am I not allowed to do that. Thanks.

Try the change of variables x -> pi/2-x to get a new integral. Then add it to the old integral.
 
Thank you =) I found the answer.
 
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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