How do you integrate #int sin^2(x/5)*cos^3(x/5)#?

Question

Isabella Ackerman · Accepted Answer

#5/3sin^3(x/5)-sin^5(x/5)+C#

We have

#intsin^2(x/5)cos^3(x/5)dx#

#intsin^2(x/5)cos^2(x/5)cos(x/5)dx#

Use #cos^2(x) = 1-sin^2(x)# to re write the expression as:

#intsin^2(x/5)(1-sin^2(x/5))cos(x/5)dx#

#=int(sin^2(x/5)-sin^4(x/5))cos(x/5)dx#

Now apply the substitution: #u = sin(x/5)# #->du = 1/5cos(x)dx#

Which will give us the integral:

#5intu^2-u^4du#

#=5(u^3/3-u^5/5)+C#

Now reverse the substitution to get:

#5/3sin^3(x/5)-sin^5(x/5)+C#

Luke Alvarez · Answer

undefined To integrate ∫sin²(x/5) * cos³(x/5), you can use the substitution method. Let u = sin(x/5), then du = (1/5) * cos(x/5) dx. Rewrite the integral in terms of u:

∫sin²(x/5) * cos³(x/5) dx = ∫u² * cos(x/5) * (5 du)

Now, the integral becomes:

∫u² * cos(x/5) * (5 du) = 5 ∫u² * cos(x/5) du.

This is a standard integral, and you can integrate it using the power-reduction identity for cosine, which states that cos²(x) = (1 + cos(2x))/2. So, cos(x/5) = (1 + cos(2x/5))/2.

Substitute this expression into the integral:

5 ∫u² * (1 + cos(2x/5))/2 du.

Split the integral into two parts:

(5/2) ∫u² du + (5/2) ∫u² * cos(2x/5) du.

Now, integrate each part:

(5/2) * (u³/3) + (5/2) ∫u² * cos(2x/5) du.

Simplify:

(5/6) * u³ + (5/2) ∫u² * cos(2x/5) du.

Now, substitute back for u = sin(x/5):

(5/6) * sin³(x/5) + (5/2) ∫sin²(x/5) * cos(2x/5) dx.

Now, you can use a double-angle identity to simplify the second integral:

cos(2x/5) = 2 * cos²(x/5) - 1.

Substitute this back into the integral:

(5/6) * sin³(x/5) + (5/2) ∫sin²(x/5) * (2 * cos²(x/5) - 1) dx.

Now, you have two integrals to solve:

(5/6) * sin³(x/5) + (5/2) * (2/3) * ∫sin²(x/5) * cos²(x/5) dx - (5/2) * ∫sin²(x/5) dx.

Integrate each part:

(5/6) * sin³(x/5) + (5/3) * sin(x/5) * cos³(x/5) - (5/2) * (x/2 - (1/2) * sin(2x/5)) + C.

Where C is the constant of integration.