Chapter 6: Techniques of Integration
Section 6.2: Trigonometric Integrals
Evaluate the indefinite integral ∫tan4xsec6x ⅆx.
As per Table 6.2.8, the given integral yields to the substitution u=tanx because the exponent on the factor sec6x is the even number 6. The complete solution appears in Table 6.2.3(a).
= ∫tan4xsec2x2sec2x ⅆx
= ∫tan4x1+tan2x2sec2x ⅆx
= ∫u41+u22 ⅆu
=19 tan9x+27 tan7x+15 tan5x
Table 6.2.3(a) Evaluation of ∫tan4xsec6x ⅆx via the substitution u=tanx
The annotated stepwise solution in Table 6.2.15 is obtained with the
tutor after both the Sum and Constant Multiple rules were selected as Understood Rules. There are two major ideas in the solution, the rewrite and the change of variables u=tanx. Left to its own devices, the tutor would make the first rewrite, but then expand 1+tan2x2and make another rewrite in which the integrand itself is completely expanded. The guided solution in Table 6.2.3(b) is a bit more compact, but follows the same strategy, namely, that of Table 6.2.8.
Table 6.2.3(b) Annotated stepwise evaluation of ∫tan4xsec6x ⅆx via the Integration Methods tutor
Note that an annotated stepwise solution is available via the Context Panel with the "All Solution Steps" option.
The rules of integration can also be applied via the Context Panel, as per the figure to the right.
Table 6.2.3(c) contains a command-based solution from first principles.
Control-drag the given integral.
2-D Math≻Convert To≻Inert Form
Context Panel: Assign to a Name≻q
∫tan4xsec6x ⅆx→assign to a nameq1
Use the eval command to rewrite the integrand.
Apply the Change command from the IntegrationTools package to effect the change of variables u=tanx.
Use the value command to evaluate the resulting integral.
Use the eval command to replace u with tanx.
Table 6.2.3(c) Command-based evaluation from first principles
Maple itself seems to take a different approach to the evaluation of the integral:
∫tan4xsec6x ⅆx = 19⁢sin⁡x5cos⁡x9+463⁢sin⁡x5cos⁡x7+8315⁢sin⁡x5cos⁡x5
That these two forms of the solution are equivalent can be established by comparing the difference:
simplify19sin⁡x5cos⁡x9+463sin⁡x5cos⁡x7+8315sin⁡x5cos⁡x5−19tan⁡x9+27tan⁡x7+15tan⁡x5 = 0
The reader who is so inclined can massage one form to the other by applying the identities sinx=cosx⋅tanx,cosx=1/secx, and sec2x=1+tan2x.
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