How do you find the integral of #int cos^3(x) sin^4(x) dx#?

Question

Evelyn Agard · Accepted Answer

#(sin^5 x)/5 + (sin^7 x)/7 +C# #intcos^3 xsin^4 xdx=intcosxcos^2 xsin^4 xdx=# #intcosx(1-sin^2 x)sin^4 xdx=I# #sinx=t => cosxdx=dt# #I=int(1-t^2)t^4dt=t^5/5-t^7/7+C=# #I=(sin^5 x)/5 + (sin^7 x)/7 +C#

Luke Alvarez · Answer

undefined To find the integral of cos^3(x) sin^4(x) dx, you can use integration by parts, where one part of the integral is differentiated and the other part is integrated. Here's how you can do it:

Let u = cos(x) and dv = sin^4(x) dx
Then, du = -sin(x) dx and v = (-1/5)cos^5(x)

Now, apply the integration by parts formula:

∫u dv = uv - ∫v du

Substitute the values of u, dv, du, and v:

∫cos^3(x) sin^4(x) dx = (-1/5)cos^5(x)cos(x) - ∫(-1/5)cos^5(x)(-sin(x)) dx

Simplify and integrate the remaining integral:

= (-1/5)cos^6(x) + (1/5)∫cos^5(x)sin(x) dx

To integrate the remaining term, you can use a substitution method or other integration techniques as necessary.

Christopher Allers · Answer

undefined To find the integral of ∫cos^3(x) sin^4(x) dx, you can use the trigonometric identity:

sin^2(x) = 1 - cos^2(x)

Substitute sin^2(x) in terms of cos^2(x), then use u-substitution where u = cos(x) and du = -sin(x) dx. After integrating, you can convert back to the original variable using the original trigonometric identities.