# How do you integrate #int cos theta * cos^5(sin theta) #?

We can try applying a substitution.

Now putting this substitution into the integral:

At this point I will quote the trig identity:

(I know this identity is a bit out of the blue. If you are having trouble convincing yourself this is true there are a few of methods to prove this but if you are good with complex numbers and binomial theorem then this question, https://tutor.hix.ai gives an example of a method that you can use to work out any of these identities. Although it is not the same identity as this one the same idea can be applied.)

Using the identity to re write the integral we get:

Doing the integration we get:

We can now reverse the substitution to get:

Hence our final answer.

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To integrate ( \int \cos(\theta) \cdot \cos^5(\sin(\theta)) ), you can use the substitution method. Let ( u = \sin(\theta) ). Then, ( du = \cos(\theta) , d\theta ). Substituting these values, the integral becomes ( \int \cos^5(u) , du ). To solve this integral, you can use the power-reducing formula for cosine raised to an odd power, which is ( \cos^n(x) = \frac{1}{2^n} \sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \binom{n}{k} \cos((n-2k)x) ). Applying this formula with ( n = 5 ), you get ( \cos^5(x) = \frac{1}{32}(10\cos(x) + 5\cos(3x) + \cos(5x)) ). Substituting ( u = \sin(\theta) ) back in, the integral becomes ( \int \frac{1}{32}(10\cos(\sin(\theta)) + 5\cos(3\sin(\theta)) + \cos(5\sin(\theta))) , d(\sin(\theta)) ). Integrate each term separately with respect to ( u ), then resubstitute ( \sin(\theta) ) back for ( u ) to obtain the final result.

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