# Thread: Problems with a few Integrals

1. ## Problems with a few Integrals

I just started Calc II and the first thing the teacher did was give us a review sheet of about 60 integrals and derivatives. There are 3 integrals in particular that I have been going around in circles with.

$=Integral 1)$cos^2[(1/2)x]dx

I know that the answer wanted is (1/2)x + (1/2)sinx + C and that the half angle formula gets me on on the correct path but I am missing something...

2) $sin^2(3x)dx Again, I have been given the answer (1/2)x - (1/12)sin(6x) + C. This gives a hint to use the double angle forumua but I'll be darned if I can make it work. 3)$sin^4(x)dx

The answer given is (1/32)[12x-8sin(2x) + sin(4x)] + C. I have tried breaking this into (sin^2(x))^2 = (1 - cos^2(x))^2 = 1 - 2cos^2(x) + cos^4(x) and again I am doing something wrong.

I am sorry for posting 3 at a time. Any help/hints will be greatly appreciated!

Quasar

2. Originally Posted by Quasar

1) $cos^2[(1/2)x]dx Use half-angle identity,$\displaystyle \int \frac{1}{2}+\frac{1}{2}\cos x dx$The rest is simple,$\displaystyle \frac{1}{2}x+\frac{1}{2}\sin x+C$2)$sin^2(3x)dx
Again half-angle identity,
$\displaystyle \int \frac{1}{2}-\frac{1}{2}\cos 6x dx$
Now,
$\displaystyle \int \frac{1}{2} dx=\frac{1}{2}x$
And,
$\displaystyle \int \frac{1}{2} \cos 6x dx = \frac{1}{12}\sin 6x$
If you use $\displaystyle u=6x$.
Thus,
$\displaystyle \frac{1}{2}x-\frac{1}{12}\sin 6x+C$

3. Originally Posted by Quasar
3) $sin^4(x)dx You have,$\displaystyle \int \sin^4 x dx\displaystyle \int (\sin^2 x)^2 dx$Use half-angle identity,$\displaystyle \int \left( \frac{1}{2}-\frac{1}{2}\cos 2x \right)^2 dx$Open sesame,$\displaystyle \int \frac{1}{4}-\frac{1}{2}\cos 2x+\frac{1}{4}\cos^2 2x dx$Use half-angle identity again,$\displaystyle \int \frac{1}{4}-\frac{1}{2}\cos 2x +\frac{1}{4}\left( \frac{1}{2}+\frac{1}{2}\cos 4x \right) dx$Open sesame,$\displaystyle \int \frac{1}{4}-\frac{1}{2}\cos 2x +\frac{1}{8}+\frac{1}{8}\cos 4x dx$Combine the brethren,$\displaystyle \int \frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x dx$Thus,$\displaystyle \frac{3}{8}x-\frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x +C$Factor,$\displaystyle \frac{1}{32} \left( 12x - 8\sin 2x+\sin 4x\right)+C\$