Chapter 6: Techniques of Integration
Section 6.2: Trigonometric Integrals
Example 6.2.5
Evaluate the indefinite integral ∫sec3x ⅆx.
Solution
Mathematical Solution
The given indefinite integral yields to a parts integration, an application of the trig identity tan2x=sec2x−1, and the device of solving for the unknown integral. In particular, take u=secx so du=secxtanx; then dv=sec2x dx so v=tanx. The complete calculation can be seen in Table 6.2.5(a).
∫sec3x ⅆx
=secxtanx−∫tanx⋅secxtanx ⅆx
=secxtanx−∫secx⋅tan2x ⅆx
=secxtanx−∫secx⋅sec2x−1 ⅆx
=secxtanx−∫sec3x ⅆx+∫secx ⅆx
=secxtanx−∫sec3x ⅆx+ln(secx+tanx)
2∫sec3x ⅆx
=secxtanx+ln(secx+tanx)
=12(secxtanx+ln(secx+tanx))
Table 6.2.5(a) Evaluation of ∫sec3x ⅆx
Maple Solution
Evaluation in Maple
Control-drag the given integral.
Context Panel: Evaluate and Display Inline
∫sec3x ⅆx = 12⁢sin⁡xcos⁡x2+12⁢ln⁡sec⁡x+tan⁡x
Note that Maple expresses secxtanx in terms of the sine and cosine functions, and that the logarithm is not taken of the absolute value of secx+tanx. Should this sum be negative, the logarithm will be complex, and a complex constant of integration would compensate.
Table 6.2.5(b) contains a solution generated with the tutor after the Sum, Constant Multiple, and Secant rules were selected as Understood Rules.
∫sec⁡x3ⅆx=tan⁡x⁢sec⁡x−∫tan⁡x2⁢sec⁡xⅆxparts,sec⁡x,tan⁡x=tan⁡x⁢sec⁡x−∫sec⁡x⁢sec⁡x2−1ⅆxrewrite,tan⁡x2=sec⁡x2−1=tan⁡x⁢sec⁡x−∫sec⁡x3ⅆx+ln⁡sec⁡x+tan⁡xrewrite,sec⁡x⁢sec⁡x2−1=sec⁡x3−sec⁡x=tan⁡x⁢sec⁡x2+ln⁡sec⁡x+tan⁡x2solve
Table 6.2.5(b) Annotated stepwise evaluation of ∫sec3x 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.
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