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What is the answer to this trig substitution problem (x^4)?
Yes, please.
Alright, the first step is to substitute x with 4sin(theta). Then, we have dx = 4cos(theta) d(theta). We can also express the square root term as 4cos(theta) since 16-x^2 = 16-16sin^2(theta) = 16cos^2(theta). So, our integral becomes: integral of 64sin^4(theta)(4cos(theta)) d(theta). Then, we can simplify further using the identity 2sin(theta)cos(theta) = sin(2theta), which gives us: integral of 32sin^4(theta)cos^2(theta) d(theta) = 16integral of (1-cos(2theta))^2(sin^2(theta)) d(theta).