Section 5.8 : Substitution Rule for Definite Integrals
7. Evaluate the following integral, if possible. If it is not possible clearly explain why it is not possible to evaluate the integral.
∫3π2π6sin(2w)−7cos(w)dwShow All Steps Hide All Steps
Start SolutionThe first step that we need to do is do the substitution.
At this point you should be fairly comfortable with substitutions. If you are not comfortable with substitutions you should go back to the substitution sections and work some problems there.
Before setting up the substitution we’ll need to break up the integral because the second term doesn’t need a substitution. Doing this gives,
∫3π2π6sin(2w)−7cos(w)dw=∫3π2π6sin(2w)dw−∫3π2π7cos(w)dwThe substitution for the first integral is then,
u=2w Show Step 2Here is the actual substitution work for this first integral.
du=2dw→dw=12duw=π:u=2πw=3π2:u=3πAs we did in the notes for this section we are also going to convert the limits to u’s to avoid having to deal with the back substitution after doing the integral.
Here is the integral after the substitution.
∫3π2π6sin(2w)−7cos(w)dw=3∫3π2πsin(u)du−∫3π2π7cos(w)dw Show Step 3The integral is then,
∫3π2π6sin(2w)−7cos(w)dw=−3cos(u)|3π2π−7sin(w)|3π2π=(3−(−3))+(7−0)=13