Calculus 2 - Trig Integrals Question (Integrating cos^2x)

1. Here's the problem on trig integrating that I'm struggling with (Calculus 2 btw)

2. Wanted to see if I did everything right so far and what to do after all this. The part where I'm stuck is how to integrate (integral)cos^(2)udu and (integral)cos^(2)usin^(2)udu. I'm sure these are easy integrals, but I'm terrible with trig identities and when or how to use them. Any suggestions?

Attachments

• 41.7 KB Views: 844

Related Calculus and Beyond Homework Help News on Phys.org
Mark44
Mentor
1. Here's the problem on trig integrating that I'm struggling with (Calculus 2 btw)
View attachment 220007
2. Wanted to see if I did everything right so far and what to do after all this. The part where I'm stuck is how to integrate (integral)cos^(2)udu and (integral)cos^(2)usin^(2)udu. I'm sure these are easy integrals, but I'm terrible with trig identities and when or how to use them. Any suggestions?
Please don't delete the homework template -- its use is required here.

##\cos^2(u) = \frac {1 + \cos(2u)} 2##
##\sin^2(u) = \frac {1 - \cos(2u)} 2##
The first one can be used directly in your first integral. For the second integral, replace each ##\cos^2(u)## factor, multiply out, and work from there, using the identities above, as needed.

Last edited:
Ray Vickson
Homework Helper
Dearly Missed
1. Here's the problem on trig integrating that I'm struggling with (Calculus 2 btw)
View attachment 220007
2. Wanted to see if I did everything right so far and what to do after all this. The part where I'm stuck is how to integrate (integral)cos^(2)udu and (integral)cos^(2)usin^(2)udu. I'm sure these are easy integrals, but I'm terrible with trig identities and when or how to use them. Any suggestions?
I have not read your work because I do not look at handwritten images---only typed work. However, it looks like you are doing a lot of extra work. It might be better to use integration by parts.

If ##I_n = \int \cos^n (u) \, du##, then integration by parts gives
$$\begin{array}{rl}I_n = &\sin(u) \cos^{n-1}(u) + (n-1) \int \sin^2(u) \cos^{n-2}(u) \, du\\ =& \sin(u) \cos^{n-1}(u) + (n-1) \int \cos^{n-2} (u) \, du - (n-1) \int \cos^n (u) \, du, \end{array}$$
and the last term is ##-(n-1) I_n##,
so we get an equation for ##I_n##:
$$I_n = \frac{1}{n} \sin(u) \cos^{n-1}(u) + \frac{n-1}{n} I_{n-2}.$$
So, you can express ##I_8## in terms of ##I_6##, ##I_6## in terms of ##I_4##, etc. That process ends at ##I_2##, which is easy.

ehild
Homework Helper
cos4(u)=[cos2(u)]2. Apply @Mark44's hint twice, then you have cos(2u) and cos(4u) and a constant to integrate.