Question Statement
Evaluate the integral:
∫03πcos2θsinθ,dθ
Background and Explanation
To solve this integral, we use a substitution method. The integral contains both cos2θ and sinθ, which suggests we can make a substitution to simplify the expression. Specifically, we’ll use a substitution to eliminate the product of trigonometric functions and solve for the integral step by step.
Solution
Start by recognizing that the integral involves a product of cos2θ and sinθ. We can use substitution to simplify the integral.
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Substitute: Let u=cosθ. Then, the derivative of u with respect to θ is du=−sinθ,dθ.
This transforms the integral as follows:
∫03πcos2θsinθ,dθ=−∫03πcos2θ(−sinθ),dθ
This simplifies to:
=−∫03πu2,du
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Change the limits: When θ=0, u=cos0=1, and when θ=3π, u=cos3π=21. So, the limits of integration change from 0 to 3π for θ, to 1 to 21 for u.
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Integrate: Now, we integrate u2 with respect to u:
∫u2,du=3u3
Thus, the integral becomes:
−[3u3]121
- Evaluate the limits: Substitute the limits into the result:
=−[3(21)3−(1)3]
Simplify the expression:
=−[381−1]=−31[81−1]
=−31[81−8]=−31×8−7
- Final Answer: Simplifying this, we get:
=247
- Substitution: Let u=cosθ, so du=−sinθ,dθ.
- Integral of a power: ∫u2,du=3u3.
Summary of Steps
- Use the substitution u=cosθ, and convert the integral.
- Change the limits of integration according to u=cosθ.
- Integrate u2 with respect to u.
- Substitute the limits into the integral and simplify.
- The final answer is:
247