# Integration problems. (Integration by parts)

1. Feb 14, 2008

### BioCore

1. The problem statement, all variables and given/known data

Hi, I have a test coming up soon so I was doing some questions from the textbook when I stumbled upon this one and I'm stuck after like 5 tries. Here is the question:

$$\int$$cos^2(x)dx

Solve.

2. Relevant equations

the question then states we should solve using this:
cos^2(x)dx = (cos x)(cos x)

which gives us:
$$\int$$cos^2(x)dx = sinxcosx + $$\int$$sin^2(x)dx

finally we should use sin^2(x) + cos^2(x) = 1 to replace the sin^2(x) at the right side of integral.

3. The attempt at a solution
so basically I tried using integral by parts, since we are studying this topic currently
and set:

u= 1 - cos^2(x) and dv = dx
du = 2cosxsinx and v = x

When I plug in the values into the integration and try to solve I don't end up with their answer. I must be overlooking something and this is where I am stuck:

$$\int$$cos^2(x)dx = sinxcosx + x(1-cos^2x) - $$\int$$2xcosxsinx

$$\int$$cos^2(x)dx = 1/2sinxcosx + 1/2x + C

Thanks for the help.

2. Feb 14, 2008

### Nesha

All problems of that type can be solved using this "formulas" :

$$\int$$cos^n(x)dx = 1/nsinxcos^n-1(x) +n-1/n $$\int$$sin^n-2(x)dx

$$\int$$sin^n(x)dx = -1/nsin^n-1(x)cosx +n-1/n $$\int$$sin^n-2(x)dx

3. Feb 14, 2008

### sutupidmath

well, you could also use this identity

cos^2(x)=(1+cos(2x))/2, it would really help you get to the answer pretty quickly, and without needing to do ineg by parts at all.

4. Feb 14, 2008

### HallsofIvy

Staff Emeritus
However, the point of the method that was suggested originally is that, after the first integration you have, as you say,
$$\int cos^2(x)dx = sinxcosx + \int sin^2(x)dx$$
Now let $sin^2(x)= 1- cos^2(x)$ and that becomes
$$\int cos^2(x)dx= sin(x) cos(x)+ \int (1- cos^2(x))dx$$
$$\int cos^2(x)dx= sin(x) cos(x)+ \int dx- \int cos^2(x) dx$$
$$\int cos^2(x)dx= sin(x) cos(x)+ x - \int cos^2(x) dx$$
Add $\int cos^2(x)dx$ to both sides of the equation and you are almost done.

Last edited: May 12, 2009
5. Feb 14, 2008

### BioCore

Halls off Ivy thanks a lot, I forgot that rule of integration. thanks a lot, yeah now that I tried it out it actually works. Thanks a lot.

6. May 12, 2009

### Shovna

$$cos^2(x)dx= 1+cos2x/2$$

wat i wanna ask is.....is this the formula for cos^2x dx or can it be used wid $$\int$$ too??