Integral of (cos x)^2 - Solution 1/2x + 1/4 sin2x

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integral (cos x)^2??
the answer is 1/2x + 1/4 sin2x
pls help...thanx...
 
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teng125 said:
integral (cos x)^2??
the answer is 1/2x + 1/4 sin2x
pls help...thanx...
To integrate something like sin2x dx, cos2x dx, we use Power-reduction formulas, ie:
\cos ^ 2 x = \frac{1 + \cos(2x)}{2} \quad \mbox{and} \quad \sin ^ 2 x = \frac{1 - \cos(2x)}{2}
Can you go from here?
 
You can do this using trig identities:
\cos{2\theta}=\cos^2{\theta}-\sin^2{\theta} = 2\cos^2{\theta}-1
Now solve for \cos^2{\theta}:
\cos^2{\theta}=\frac{1+\cos{2\theta}}{2}
So
\int\cos^2{x}dx=\frac{1}{2}\int 1+\cos{2x}dx
 
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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