What is the integral of #sin^2(x)cos^4(x) #?

Question

Charles Arocha · Accepted Answer

#1/192(12x+3sin2x-3sin4x-sin6x)+C#.

Let #I=intsin^2xcos^4xdx#. We will use the following Identities to simplify the Integrand :- # [1] :2sin^2theta=1-cos2theta, [2] : 2cos^2theta=1+cos2theta# # [3] : 2cosCcosD=cos(C+D)+cos(C-D)# Now, #sin^2xcos^4x=1/8(4sin^2xcos^2x)(2cos^2x)# #=1/8(2sinxcosx)^2(1+cos2x)# #=1/8(sin2x)^2(1+cos2x)# #=1/8(sin^2 2x)(1+cos2x)# #=1/16(2sin^2 2x)(1+cos2x)# #=1/16(1-cos4x)(1+cos2x)# #=1/16(1-cos4x+cos2x-cos4xcos2x)# #=1/16{1-cos4x+cos2x-1/2(cos6x+cos2x)}# #=1/32(2+cos2x-2cos4x-cos6x)# #:. I=1/32int(2+cos2x-2cos4x-cos6x)dx# #=1/32(2x+sin(2x)/2-(2sin(4x))/4-sin(6x)/6)# #=1/192(12x+3sin2x-3sin4x-sin6x)+C#. Enjoy Maths.!

Cameron Alexander · Answer

undefined The integral of $\sin^2(x)\cos^4(x)$ can be computed using trigonometric identities and integration techniques. Here's how to do it:

1. Use the identity $\sin^2(x) = 1 - \cos^2(x)$ to rewrite the integral as $(1 - \cos^2(x))\cos^4(x)$.
2. Expand the expression to get $\cos^4(x) - \cos^6(x)$.
3. Split the integral into two parts: $\int \cos^4(x) \, dx$ and $\int \cos^6(x) \, dx$.
4. For $\int \cos^4(x) \, dx$, use the reduction formula $\int \cos^n(x) \, dx = \frac{\cos^{n-1}(x)\sin(x)}{n} + \frac{n-1}{n}\int \cos^{n-2}(x) \, dx$.
5. For $\int \cos^6(x) \, dx$, use a similar reduction formula.
6. After integrating both parts, combine the results to find the integral of $\sin^2(x)\cos^4(x)$.