How Do You Approach Tricky Integration Problems Using Trig Identities?

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Hello, I'm new to the forums. I've lurked for awhile but I've decided to join up. Anyways, my math professor assigned homework over the break on material he didn't cover in class and is not in our textbook. Any help would be appreciated...

Ok, we're supposed to use basic trig identities to get the integral down to something manageable where we can use u-substitution and/or integration by parts. I've worked out one but I'm having trouble on the other two...

1. the integral of (cos(x))^4 dx. [Hint: also (cos(x)^2)^2]

2. the integral of sec(v) dv. [Hint: multiply sec(v) by (sec(v)+tan(v))/(sec(v)+tan(v)).]

Some identities to use:

sin(2x)=2sin(x)cos(x)

cos(2x)=(cos(x))^2-(sin(x))^2

cos(2x)=2(cos(x))^2-1

cos(2x)=1-2(sin(x))^2

(sin(x))^2=(1/2)[1-COS(2X)]

I'm not looking for the answer, just a push in the right direction, thanks...
 
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Some other useful trigonometric identities:

\sin^2\frac{x}{2}=\frac{1-\cos x}{2},
\cos^2\frac{x}{2}=\frac{1+\cos x}{2}.

Edit: actually, these are pretty much the same as your last identity. o:)
 
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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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