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Solve each of the definite integrals in Exercises 67–76.

∫0πsin3xcos2xdx.

Short Answer

Expert verified

The answer is∫0πsin3xcos2xdx=415

Step by step solution

01

Step 1. Given Information

The integral is∫0πsin3xcos2xdx.

02

Step 2. Explanation

Rewrite the integral and substitute sin2x=1-cos2x.

∫0πsin2xcos2xsinxdx=∫0π1-cos2xcos2xsinxdx

Substitute localid="1649142933889" u=cosx. Differentiate it.

localid="1649143357676" du=-sinxdx.

Limit changes to -1 to 1. so, the integral becomes,

∫1-1-(1-u2)u2du=∫1-1-u2+u4du

Change the limit.

-∫-11-u2+u4du

03

Step 3. Calculation

Solve the integral.

-∫-11-u2du+∫-11u4du=--∫-11u2du+∫-11u4du=--u33-11+u55-11=--23+25=--415=415

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