Chapter 6: Techniques of Integration
Section 6.2: Trigonometric Integrals
Example 6.2.7
Derive the first reduction formula in Table 6.2.1.
Solution
Mathematical Solution
The derivation in Table 6.2.7(a) begins with an integration by parts in which
u=cosm−1x
dv=cosxsinnxdx
du=m−1cosm−2x−sinxdx
v=sinn+1xn+1
This creates a factor of sinn+2x in the new integral on the right, a factor that is replaced by 1−cos2x. 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+1xn+1−−m−1n+1 ∫cosm−2xsinn+2x dx
=cosm−1xsinn+1xn+1+m−1n+1 ∫cosm−2xsinnx1−cos2x dx
=cosm−1xsinn+1xn+1+m−1n+1 ∫cosm−2xsinnx dx−∫cosmxsinnx dx
=cosm−1xsinn+1xn+1+m−1n+1 ∫cosm−2xsinnx dx−m−1n+1 ∫cosmxsinnx dx
1+m−1n+1∫cosmxsinnx ⅆx
=cosm−1xsinn+1xn+1+m−1n+1 ∫cosm−2xsinnx dx
=n+1m+ncosm−1xsinn+1xn+1+m−1n+1∫cosm−2xsinnx dx
=cosm−1xsinn+1xm+n+m−1m+n∫cosm−2xsinnx dx
Table 6.2.7(a) Derivation of the first 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,cosm−1x
sin⁡xn+1⁢cos⁡xm−1n+1−∫−sin⁡xn+1⁢cos⁡xm−1⁢m−1⁢sin⁡xn+1⁢cos⁡xⅆx
Massage to the form in line 1 of Table 6.2.7(a)
q2≔collectsimplifyq1,int
sin⁡xn+1⁢cos⁡xm−1n+1+m−1⁢∫sin⁡xn+2⁢cos⁡xm−2ⅆxn+1
Replace sin2x with 1−cos2x and massage to form in line 4 of Table 6.2.7(a)
q3≔collectsimplifyExpandevalq2,sinn+2x=sinnx⋅1−cos2x,int
−m+1⁢∫cos⁡xm⁢sin⁡xnⅆxn+1+m−1⁢∫sin⁡xn⁢cos⁡xm−2ⅆxn+1+sin⁡xn+1⁢cos⁡xm−1n+1
Solve for the unknown integral
collectnormalisolateQ=q3,Q,int
∫cos⁡xm⁢sin⁡xnⅆx=m−1⁢∫sin⁡xn⁢cos⁡xm−2ⅆxm+n+sin⁡xn+1⁢cos⁡xm−1m+n
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