Solve the Trig Integral. ∫sin^2(x)cos^4(x) dx

Question

Solve the Trig Integral. Show all steps

∫sin²(x)cos⁴(x) dx

Answer 1

let's try

(sin²x)(cos²x)=(1/4)sin²2x

cos²x=(1/2)(1+cos2x)

now integrand is [(1/8)(sin²2x+sin²2x(cos2x))]dx

and for the first sin²2x, sin²2x=(1-cos4x)/2

integrand is [(1/16)-(1/16)cos4x+(sin²2x)(cos2x)/8]dx

integrating\

I=(x/16)-(1/64)sin4x+(1/48)sin³2x+C

Answer 2

We know that sin²(x) + cos²(x) = 1, so sin²(x) = 1 - cos²(x).

We now need to evaluate ∫((cos⁴(x)) * (1 - cos²(x)))dx = ∫(cos⁴(x) - cos⁶(x))dx

If we want to solve ∫(cos⁴(x))dx, we see that this is equivalent to solving ∫((cos²(x))²dx and cos²(x) = (1 + cos(2x))/2. Substitute that in, and we see that we need to solve

∫(1/2 + cos(2x)/2)²dx = 1/4 * ∫(1 + 2cos(2x) + cos²(2x))dx. We know that cos²(2x) = (1 + cos(4x)) / 2. So, we just have to solve 1/4 * ∫(1 + 2cos(2x) + 1/2 + cos(4x)/2)dx which can be done by u-substitution

If we want to sovle ∫cos⁶(x)dx, we can frame the problem as solving ∫(cos²(x))³dx. Using the earlier fact about cos²(x)

∫(cos²(x))³dx = ∫(1/2 + cos(2x)/2)³dx. We can then expand out this expression and solve each of the individual pieces in separate integrals. Feel free to let me know if you have further questions! Hope this helps :)