Chapter 6: Techniques of Integration
Section 6.2: Trigonometric Integrals
Example 6.2.8
Derive the second reduction formula in Table 6.2.1.
Solution
Mathematical Solution
The derivation in Table 6.2.8(a) begins with an integration by parts in which
u=sinn−1x
dv=sinxcosmxdx
du=n−1sinn−2xcosxdx
v=−cosm+1xm+1
This creates a factor of cosm+2x in the new integral on the right, a factor that is replaced by 1−sin2x. This replacement then creates an integral on the right that is a multiple of the original integral on the left. Solving for the unknown integral results in the desired reduction formula.
∫cosmxsinnx ⅆx
=−cosm+1xsinn−1xm+1−−n−1m+1 ∫cosm+2xsinn−2x dx
=−cosm+1xsinn−1xm+1+n−1m+1 ∫cosmxsinn−2x1−sin2x dx
=−cosm+1xsinn−1xm+1+n−1m+1 ∫cosmxsinn−2x dx−∫cosmxsinnx dx
=−cosm+1xsinn−1xm+1+n−1m+1 ∫cosmxsinn−2x dx−n−1m+1 ∫cosmxsinnx dx
1+n−1m+1∫cosmxsinnx ⅆx
=−cosm+1xsinn−1xm+1+n−1m+1 ∫cosmxsinn−2x dx
=m+1m+n−cosm+1xsinn−1xm+1+n−1m+1∫cosmxsinn−2x dx
=−cosm+1xsinn−1xm+n+n−1m+n∫cosmxsinn−2x dx
Table 6.2.8(a) Derivation of the second reduction formula in Table 6.2.1
Maple Solution
Initialize
Install the IntegrationTools package.
withIntegrationTools:
Assign the name Q to the given integral.
Q≔∫cosmxsinnx ⅆx:
Integrate by parts
q1≔PartsQ,sinn−1x
−cos⁡xm+1⁢sin⁡xn−1m+1−∫−cos⁡xm+1⁢sin⁡xn−1⁢n−1⁢cos⁡xm+1⁢sin⁡xⅆx
Massage to the form in line 1 of Table 6.2.8(a)
q2≔collectsimplifyq1,int
−−n+1⁢∫cos⁡xm+2⁢sin⁡xn−2ⅆxm+1−cos⁡xm+1⁢sin⁡xn−1m+1
Replace cos2x with 1−sin2x and massage to form in line 4 of Table 6.2.8(a)
q3≔collectsimplifyExpandevalq2,cosm+2x=cosmx⋅1−sin2x,int
−n+1⁢∫cos⁡xm⁢sin⁡xnⅆxm+1+n−1⁢∫cos⁡xm⁢sin⁡xn−2ⅆxm+1−cos⁡xm+1⁢sin⁡xn−1m+1
Solve for the unknown integral
collectnormalisolateQ=q3,Q,int
∫cos⁡xm⁢sin⁡xnⅆx=n−1⁢∫cos⁡xm⁢sin⁡xn−2ⅆxm+n−sin⁡xn−1⁢cos⁡xm+1m+n
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