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

[teng125]

integral (cos x)^2??
the answer is 1/2x + 1/4 sin2x
pls help...thanx...

 

[VietDao29]

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?

 

[LeonhardEuler]

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
$$\int\cos^2{x}dx=\frac{1}{2}\int 1+\cos{2x}dx$$