Integration problems. (Integration by parts)

[BioCore]

Homework Statement

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:

\intcos^2(x)dx

Solve.

Homework Equations

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

which gives us:
\intcos^2(x)dx = sinxcosx + \intsin^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.

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:

\intcos^2(x)dx = sinxcosx + x(1-cos^2x) - \int2xcosxsinx

The final answer should be:
\intcos^2(x)dx = 1/2sinxcosx + 1/2x + C

Thanks for the help.

 

[Nesha]

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

\intcos^n(x)dx = 1/nsinxcos^n-1(x) +n-1/n \intsin^n-2(x)dx

\intsin^n(x)dx = -1/nsin^n-1(x)cosx +n-1/n \intsin^n-2(x)dx

 

[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.

 

[HallsofIvy]

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.

 

[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.

 

[Shovna]

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




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