Math  /  Calculus

QuestionUse a suitable change of variables to determine the indefinite integral. (1cos4(x))5cos3(x)sin(x)dx=+C\int\left(1-\cos ^{4}(x)\right)^{5} \cos ^{3}(x) \sin (x) d x=\square+C
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Studdy Solution

STEP 1

1. The integral involves a trigonometric expression that can be simplified using substitution.
2. We will use a substitution method to simplify the integration process.

STEP 2

1. Identify a suitable substitution.
2. Perform the substitution.
3. Integrate the resulting expression.
4. Substitute back to the original variable.

STEP 3

Identify a suitable substitution. Notice that cos(x) \cos(x) and sin(x) \sin(x) are present in the expression. We can use the substitution u=cos(x) u = \cos(x) , which implies du=sin(x)dx du = -\sin(x) \, dx .

STEP 4

Perform the substitution. Rewrite the integral in terms of u u :
Given u=cos(x) u = \cos(x) , then du=sin(x)dx du = -\sin(x) \, dx , or du=sin(x)dx -du = \sin(x) \, dx .
Substitute into the integral:
(1cos4(x))5cos3(x)sin(x)dx \int (1 - \cos^4(x))^5 \cos^3(x) \sin(x) \, dx
=(1u4)5u3(du) = \int (1 - u^4)^5 u^3 (-du)
=(1u4)5u3du = -\int (1 - u^4)^5 u^3 \, du

STEP 5

Integrate the resulting expression. The integral becomes:
(1u4)5u3du -\int (1 - u^4)^5 u^3 \, du
This requires expanding (1u4)5 (1 - u^4)^5 and integrating term by term. However, the substitution was chosen to simplify the integration, so let's reconsider the substitution or simplify further.
Re-evaluate the substitution: Let v=1cos4(x) v = 1 - \cos^4(x) , then dv=4cos3(x)sin(x)dx dv = -4\cos^3(x)\sin(x) \, dx .
Adjust the substitution:
v514dv \int v^5 \frac{-1}{4} dv
=14v5dv = -\frac{1}{4} \int v^5 \, dv

STEP 6

Integrate the simplified expression:
14v5dv -\frac{1}{4} \int v^5 \, dv
=14v66+C = -\frac{1}{4} \cdot \frac{v^6}{6} + C
=124v6+C = -\frac{1}{24} v^6 + C
Substitute back to the original variable:
v=1cos4(x) v = 1 - \cos^4(x)
=124(1cos4(x))6+C = -\frac{1}{24} (1 - \cos^4(x))^6 + C
The indefinite integral is:
124(1cos4(x))6+C \boxed{-\frac{1}{24} (1 - \cos^4(x))^6 + C}

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