Chapter 6: Techniques of Integration
Section 6.2: Trigonometric Integrals
Evaluate the indefinite integral ∫tan3x ⅆx.
The second formula in Table 6.2.8 suggests an appropriate approach: write the integrand as tanx⋅tan2x and apply the trig identity tan2x=sec2x−1. The complete calculation appears in Table 6.2.4(a).
= ∫tanx⋅tan2x ⅆx
= ∫tanxsec2x−1 ⅆx
= ∫tanxsec2x ⅆx−∫tanx ⅆx
= ∫u ⅆu−∫tanx ⅆx
Table 6.2.4(a) Indefinite integral of tan3x
Evaluation in Maple
Control-drag the given integral.
Context Panel: Evaluate and Display Inline
∫tan3x ⅆx = 12⁢tan⁡x2−12⁢ln⁡1+tan⁡x2
Note Maple's evaluation of ∫tanx dx to ln1+tan2x/2=lnsec2x/2 instead of ln(secx).
tutor, with the Sum, Constant Multiple, and Tangent rules selected as Understood Rules, gives the result shown in Table 6.2.4(b).
Table 6.2.4(b) Evaluation of ∫tan3x dx via the Integration Methods tutor
After the application of the trig identity tan2x=sec2x−1, an immediate application of the Sum rule changes the integral to
∫tanxsec2x dx−∫tanx dx
Here, Maple integrates tanx to −lncosx rather than to lnsecx, or to lnsec2x/2 as it does outside the confines of the Tangent rule of the Student Calculus1 package. In the other integral, Maple makes the change of variables u=secx rather than u=tanx, as per the shading in Table 6.2.4(b). Hence, the antiderivative is sec2x/2 rather than tan2x/2. These two antiderivatives are "equivalent" in the sense that they differ by just the additive constant of 1/2 because of the trig identity 1+tan2x=sec2x.
Table 6.2.4(c) contains the much more detailed annotated stepwise solution generated without the use of any Understood Rules.
Tools≻Load Package: Student Calculus 1
Expression palette: Indefinite-integral template
Context Panel: Student Calculus1≻All Solution Steps
∫tan3x ⅆx→show solution stepsIntegration Steps∫tan⁡x3ⅆx▫1. Rewrite◦Equivalent expressiontan⁡x2=sec⁡x2−1This gives:∫sec⁡x2−1⁢tan⁡xⅆx▫2. Rewrite◦Equivalent expressionsec⁡x2−1⁢tan⁡x=sec⁡x2⁢tan⁡x−tan⁡xThis gives:∫sec⁡x2⁢tan⁡x−tan⁡xⅆx▫3. Apply the sum rule◦Recall the definition of the sum rule∫f⁡x+g⁡xⅆx=∫f⁡xⅆx+∫g⁡xⅆxf⁡x=sec⁡x2⁢tan⁡xg⁡x=−tan⁡xThis gives:∫sec⁡x2⁢tan⁡xⅆx+∫−tan⁡xⅆx▫4. Apply a change of variables to rewrite the integral in terms of u◦Let u beu=sec⁡x◦Isolate equation for xx=arcsec⁡u◦Differentiate both sides=u2⁢1−1u2◦Substitute the values for x and dx back into the original∫sec⁡x2⁢tan⁡xⅆx=∫uⅆuThis gives:∫uⅆu+∫−tan⁡xⅆx▫5. Apply the power rule to the term ∫uⅆu◦Recall the definition of the power rule, for n ≠ -1∫uⅆu=◦This means:∫uⅆu=◦So,∫uⅆu=u22We can rewrite the integral as:u22+∫−tan⁡xⅆx▫6. Revert change of variable◦Variable we defined in step 4u=sec⁡xThis gives:sec⁡x22+∫−tan⁡xⅆx▫7. Apply the constant multiple rule to the term ∫−tan⁡xⅆx◦Recall the definition of the constant multiple rule∫⁢f⁡xⅆx=⁢∫f⁡xⅆx◦This means:∫−tan⁡xⅆx=−∫tan⁡xⅆxWe can rewrite the integral as:sec⁡x22−∫tan⁡xⅆx▫8. Rewrite◦Equivalent expressiontan⁡x=sin⁡xcos⁡xThis gives:sec⁡x22−∫sin⁡xcos⁡xⅆx▫9. Apply a change of variables to rewrite the integral in terms of u◦Let u beu=cos⁡x◦Isolate equation for xx=arccos⁡u◦Differentiate both sides=−−u2+1◦Substitute the values for x and dx back into the original∫sin⁡xcos⁡xⅆx=∫−1uⅆuThis gives:sec⁡x22−∫−1uⅆu▫10. Apply the constant multiple rule to the term ∫−1uⅆu◦Recall the definition of the constant multiple rule∫⁢f⁡uⅆu=⁢∫f⁡uⅆu◦This means:∫−1uⅆu=−∫1uⅆuWe can rewrite the integral as:sec⁡x22+∫1uⅆu▫11. Apply the reciprocal rule to the term ∫1uⅆu◦Recall the definition of the reciprocal rule∫1uⅆu=ln⁡uWe can rewrite the integral as:sec⁡x22+ln⁡u▫12. Revert change of variable◦Variable we defined in step 9u=cos⁡xThis gives:sec⁡x22+ln⁡cos⁡x
Table 6.2.4(c) Detailed annotated stepwise evaluation of ∫tan3x ⅆx
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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