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

[Civil_Disobedient]

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?

 

[Mark44]

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.

Here are two identities that will help you:
$\cos^2(u) = \frac {1 + \cos(2u)} 2$
$\sin^2(u) = \frac {1 - \cos(2u)} 2$
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.

 

[Ray Vickson]

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]

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