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Math  /  Calculus

Question∫cos⁡4θsin⁡3θdθ\int \cos ^{4} \theta \sin ^{3} \theta d \theta

Studdy Solution

STEP 1

1. We are given the integral ∫cos⁡4θsin⁡3θ dθ\int \cos^4 \theta \sin^3 \theta \, d\theta.
2. We will use trigonometric identities and substitution to simplify the integral.
3. We will aim to express the integral in terms of a single trigonometric function.

STEP 2

1. Simplify the integral using trigonometric identities.
2. Use substitution to solve the integral.
3. Integrate the resulting expression.
4. Simplify the final result.

STEP 3

First, express sin⁡3θ\sin^3 \theta in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta:
sin⁡3θ=sin⁡2θ⋅sin⁡θ=(1−cos⁡2θ)sin⁡θ\sin^3 \theta = \sin^2 \theta \cdot \sin \theta = (1 - \cos^2 \theta) \sin \theta
Substitute this into the integral:
∫cos⁡4θsin⁡3θ dθ=∫cos⁡4θ(1−cos⁡2θ)sin⁡θ dθ\int \cos^4 \theta \sin^3 \theta \, d\theta = \int \cos^4 \theta (1 - \cos^2 \theta) \sin \theta \, d\theta

STEP 4

Distribute cos⁡4θ\cos^4 \theta in the integral:
∫cos⁡4θsin⁡θ dθ−∫cos⁡6θsin⁡θ dθ\int \cos^4 \theta \sin \theta \, d\theta - \int \cos^6 \theta \sin \theta \, d\theta

STEP 5

Use substitution for each integral. Let u=cos⁡θ u = \cos \theta , then du=−sin⁡θ dθ du = -\sin \theta \, d\theta .
For the first integral:
∫cos⁡4θsin⁡θ dθ=−∫u4 du\int \cos^4 \theta \sin \theta \, d\theta = -\int u^4 \, du
For the second integral:
∫cos⁡6θsin⁡θ dθ=−∫u6 du\int \cos^6 \theta \sin \theta \, d\theta = -\int u^6 \, du

STEP 6

Integrate each expression:
For −∫u4 du-\int u^4 \, du:
−u55+C1-\frac{u^5}{5} + C_1
For −∫u6 du-\int u^6 \, du:
−u77+C2-\frac{u^7}{7} + C_2

STEP 7

Substitute back u=cos⁡θ u = \cos \theta into the integrated expressions:
−cos⁡5θ5+C1−(−cos⁡7θ7+C2)-\frac{\cos^5 \theta}{5} + C_1 - \left(-\frac{\cos^7 \theta}{7} + C_2\right)
Combine the constants:
−cos⁡5θ5+cos⁡7θ7+C-\frac{\cos^5 \theta}{5} + \frac{\cos^7 \theta}{7} + C
The final result of the integral is:
−cos⁡5θ5+cos⁡7θ7+C\boxed{-\frac{\cos^5 \theta}{5} + \frac{\cos^7 \theta}{7} + C}

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