Solve integration problems involving products and powers of tanx and secx
Use reduction formulas to solve trigonometric integrals
Before discussing the integration of products and powers of tanx and secx, it is useful to recall the integrals involving tanx and secx we have already learned:
∫sec2xdx=tanx+C
∫secxtanxdx=secx+C
∫tanxdx=ln|secx|+C
∫secxdx=ln|secx+tanx|+C.
For most integrals of products and powers of tanx and secx, we rewrite the expression we wish to integrate as the sum or difference of integrals of the form ∫tanjxsec2xdx or ∫secjxtanxdx. As we see in the following example, we can evaluate these new integrals by using u-substitution. Before doing so, it is useful to note how the Pythagorean Identity implies relationships between other pairs of trigonometric functions.
Recall: The Pythagorean Identity
For any angle x:
sin2x+cos2x=1
Dividing the original equation by cos2x and simplifying yields an expression for sec2x in terms of tan2x:
tan2x+1=sec2x
Subtracting both sides of the equation by 1 yields an expression for tan2x in terms of sec2x:
tan2x=sec2x−1
Example: Evaluating ∫secjxtanxdx
Evaluate ∫sec5xtanxdx.
Show Solution
Start by rewriting sec5xtanx as sec4xsecxtanx.
∫sec5xtanxdx=∫sec4xsecxtanxdxLet u=secx;then,du=secxtanxdx.=∫u4duEvaluate the integral.=15u5+CSubstitute secx=u.=15sec5x+C
To integrate ∫tankxsecjxdx, use the following strategies:
If j is even and j≥2, rewrite secjx=secj−2xsec2x and use sec2x=tan2x+1 to rewrite secj−2x in terms of tanx. Let u=tanx and du=sec2x.
If k is odd and j≥1, rewrite tankxsecjx=tank−1xsecj−1xsecxtanx and use tan2x=sec2x−1 to rewrite tank−1x in terms of secx. Let u=secx and du=secxtanxdx. (Note: If j is even and k is odd, then either strategy 1 or strategy 2 may be used.)
If k is odd where k≥3 and j=0, rewrite tankx=tank−2xtan2x=tank−2x(sec2x−1)=tank−2xsec2x−tank−2x. It may be necessary to repeat this process on the tank−2x term.
If k is even and j is odd, then use tan2x=sec2x−1 to express tankx in terms of secx. Use integration by parts to integrate odd powers of secx.
Example: Integrating ∫tankxsecjxdx when j is Even
Evaluate ∫tan6xsec4xdx.
Show Solution
Since the power on secx is even, rewrite sec4x=sec2xsec2x and use sec2x=tan2x+1 to rewrite the first sec2x in terms of tanx. Thus,
∫tan6xsec4xdx=∫tan6x(tan2x+1)sec2xdxLet u=tanx and du=sec2x.=∫u6(u2+1)duExpand.=∫(u8+u6)duEvaluate the integral.=19u9+17u7+CSubstitute tanx=u.=19tan9x+17tan7x+C.
Example: Integrating ∫tankxsecjxdx when k is Odd
Evaluate ∫tan5xsec3xdx.
Show Solution
Since the power on tanx is odd, begin by rewriting tan5xsec3x=tan4xsec2xsecxtanx. Thus,
Since the integral ∫sec3xdx has reappeared on the right-hand side, we can solve for ∫sec3xdx by adding it to both sides. In doing so, we obtain
2∫sec3xdx=secxtanx+ln|secx+tanx|.
Dividing by 2, we arrive at
∫sec3xdx=12secxtanx+12ln|secx+tanx|+C.
try it
Evaluate ∫tan3xsec7xdx.
Hint
Use the previous example as a guide: Integrating ∫tankxsecjxdx when k is Odd.
Show Solution
19sec9x−17sec7x+C
Watch the following video to see the worked solution to the above Try It
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
Evaluating ∫secnxdx for values of n where n is odd requires integration by parts. In addition, we must also know the value of ∫secn−2xdx to evaluate ∫secnxdx. The evaluation of ∫tannxdx also requires being able to integrate ∫tann−2xdx. To make the process easier, we can derive and apply the following power reduction formulas. These rules allow us to replace the integral of a power of secx or tanx with the integral of a lower power of secx or tanx.
Rule: Reduction Formulas for ∫secnxdx and ∫tannxdx
∫secnxdx=1n−1secn−2xtanx+n−2n−1∫secn−2xdx
∫tannxdx=1n−1tann−1x−∫tann−2xdx
The first power reduction rule may be verified by applying integration by parts. The second may be verified by following the strategy outlined for integrating odd powers of tanx.
Example: Revisiting ∫sec3xdx
Apply a reduction formula to evaluate ∫sec3xdx.
Show Solution
By applying the first reduction formula, we obtain
Applying the reduction formula for ∫tan4xdx we have
∫tan4xdx=13tan3x−∫tan2xdx=13tan3x−(tanx−∫tan0xdx)Apply the reduction formula to∫tan2xdx.=13tan3x−tanx+∫1dxSimplify.=13tan3x−tanx+x+C.Evaluate∫1dx.
try it
Apply the reduction formula to ∫sec5xdx.
Hint
Use reduction formula 1 and let n=5.
Show Solution
∫sec5xdx=14sec3xtanx−34∫sec3x
Watch the following video to see the worked solution to the above Try It
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.