Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

xcosx2sinx2dx

Short Answer

Expert verified

The solution of the given integral is xcosx2sinx2dx=sinx2+C.

Step by step solution

01

Step 1. Given Information 

Solving the given integrals.

xcosx2sinx2dx

02

Step 2. Using the substitution method.

Let

u=sinx2dudx=ddxsinx2dudx=2xcosx2du=2xcosx2dx12du=xcosx2dx
03

Step 3. This substitution changes the integral into 

xcosx2sinx2dx=121uduxcosx2sinx2dx=121u1/2duxcosx2sinx2dx=12u-1/2duxcosx2sinx2dx=12u-1/2+1-1/2+1+Cxcosx2sinx2dx=12u1/21/2+Cxcosx2sinx2dx=12·2u+Cxcosx2sinx2dx=u+Cxcosx2sinx2dx=sinx2+C

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