Integration using Trig Substitution

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sfora71

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integrate x^2(9+4x^2)^(-1/2)

My work so far:

2x=3 tan??
x=3/2 tan??
dx=3/2 (sec??)^2 d?
?(9+4x^2 )=3sec??


=?[9/4 (tan?)^2 (3/2 (sec?)^2 )]d? / (3 sec?? )
=9/8 ?(tan??)^2 sec?? d?
=9/8 ?[(sec?)^2-1]sec?? d?
=9/8(?(sec??)^3 d? - ?sec?? d?)
=9/8(sec?? tan?? - ?sec?? (tan??)^2 d?
=9/8(sec?? tan?? - ?sec?? (tan?)^2 d?)
=9/8(sec?? tan?? - ?[(sec??)^2-1)sec?? d?)
=9/8[sec?? tan?? - (?(sec?)^3 d? - ?sec?? d?)]

What should I do next?
 
sfora71 said:
integrate x^2(9+4x^2)^(-1/2)

My work so far:

2x=3 tan??
x=3/2 tan??
dx=3/2 (sec??)^2 d?
?(9+4x^2 )=3sec??


=?[9/4 (tan?)^2 (3/2 (sec?)^2 )]d? / (3 sec?? )
=9/8 ?(tan??)^2 sec?? d?
=9/8 ?[(sec?)^2-1]sec?? d?

=9/8[tan? - ln|sec? + tan?|] + C

=9/8(?(sec??)^3 d? - ?sec?? d?)
=9/8(sec?? tan?? - ?sec?? (tan??)^2 d?
=9/8(sec?? tan?? - ?sec?? (tan?)^2 d?)
=9/8(sec?? tan?? - ?[(sec??)^2-1)sec?? d?)
=9/8[sec?? tan?? - (?(sec?)^3 d? - ?sec?? d?)]

What should I do next?
Click to expand...
 
\(\displaystyle \int\frac{x^{2}}{(9+4x^{2})^{1/2}}dx \ = \ \int\frac{x^{2}}{[(3)^{2}+(2x)^{2}]^{1/2}}dx\)

\(\displaystyle Letting \ x \ = \ \frac{3tan(\theta)}{2} \ \implies \ x^{2} \ = \ \frac{9tan^{2}(\theta)}{4} \ and \ dx \ = \ \frac{3sec^{2}(\theta)d\theta}{2}\)

\(\displaystyle Now, \ this \ all \ reduces \ to \ \frac{9}{8}\int tan^{2}(\theta)sec(\theta)d\theta \ = \ \frac{9}{8} \int[sec^{2}(\theta)-1]sec(\theta)d\theta\)

\(\displaystyle You \ should \ be \ able \ to \ take \ it \ from \ here.\)
 
And to Subhotosh Khan:

the change you made is the solution for the integral of (sec??)^2 - sec??, not (sec??)^3 - sec??
 
\(\displaystyle Continuing: \ \frac{9}{8}\int sec^{3}(\theta)d\theta \ -\frac{9}{8} \int sec(\theta)d\theta\)

\(\displaystyle = \ \frac{9}{8}\bigg[\frac{sec(\theta)tan(\theta)}{2}+ \ \frac{1}{2} \int sec(\theta)d\theta\bigg] \ - \frac{9}{8}\int sec(\theta)d\theta\)

\(\displaystyle \ = \ \frac{9}{16}sec(\theta)tan(\theta)+\frac{9}{16} \int sec(\theta)d\theta-\frac{9}{8} \int sec(\theta)d\theta\)

\(\displaystyle = \ \frac{9}{16}sec(\theta)tan(\theta)- \ \frac{9}{16} \int sec(\theta)d\theta\)

\(\displaystyle = \ \frac{9}{16}\bigg[\frac{2x \sqrt{9+4x^{2}}}{9}\bigg]-\frac{9}{16}ln\bigg|\frac{2x+\sqrt{9+4x^{2}}}{3}\bigg|+C_1\)

\(\displaystyle = \ \frac{x\sqrt{9+4x^{2}}}{8}-\frac{9}{16}\bigg[ln|2x+\sqrt{9+4x^{2}}|-ln(3)\bigg]+C_1\)

\(\displaystyle = \ \frac{x\sqrt{9+4x^{2}}}{8}-\frac{9}{16}ln|2x+\sqrt{9+4x^{2}}|+\frac{9}{16}ln(3)+C_1, \ Let \ C \ = \ C_1+\frac{9}{16}ln(3), \ hence,\)

\(\displaystyle \int\frac{x^{2}}{\sqrt{9+4x^{2}}}}dx \ = \ \frac{x\sqrt{9+4x^{2}}}{8}-\frac{9}{16}ln|2x+\sqrt{9+4x^{2}}|+C\)
 
Hello, sfora71!

\(\displaystyle \text{integrate:}\;I \;=\;\int \frac{x^2\,dx}{(9+4x^2)^{\frac{1}{2}}}\)


\(\displaystyle \text{My work so far: }\;2x\:=\:3\tan\theta \quad\Rightarrow\quad x \:=\:\tfrac{3}{2}\tan\theta\)

. . \(\displaystyle dx \:=\:\tfrac{3}{2}\sec^2\!\theta\,d\theta \quad\Rightarrow\quad \sqrt{9+4x^2} \:=\:3\sec\theta\)


\(\displaystyle \text{Substitute: }\;I \;=\;\int \frac{\frac{9}{4}\tan^2\!\theta\leeft(\frac{3}{2}\sec^2\!\theta\,d\theta)}{3\sec\theta} \;=\; \tfrac{9}{8}\int\sec\theta\tan^2\!\theta\,d\theta\)

. . \(\displaystyle I \;=\;\tfrac{9}{8}\int(\sec^2\!\theta-1)\sec\theta\,d\theta \;=\;\tfrac{9}{8}\int(\sec^3\theta - \sec\theta)\,d\theta\)

. . \(\displaystyle I \;=\;\tfrac{9}{8}\int\sec^3\!\theta\,d\theta - \tfrac{9}{8}\int\sec\theta\,d\theta\)

\(\displaystyle \text{What should I do next?}\)
Click to expand...

\(\displaystyle \text{You know the formula for the second integral: }\;\boxed{\int\sec\theta\,d\theta \;=\;\ln|\sec\theta + \tan\theta| + C}\)


\(\displaystyle \text{For }\int\sec^3\!\theta\,d\theta\:\text{ we perform some Olympic-level gymnastics . . .}\)

\(\displaystyle \text{Let }\:J \:=\:\int\sec^3\!\theta\,d\theta\)

\(\displaystyle \text{Integrate by parts: }\;\begin{Bmatrix}u &=& \sec\theta && dv &=& \sec^2\!\theta\,d\theta} \\ du &=& \sec\theta\tan\theta\,d\theta && v &=& \tan\theta \end{Bmatrix}\)

\(\displaystyle \text{Then: }\;J \;=\;\sec\theta\tan\theta - \int\sec\theta\tan^2\!\theta\,d\theta\)

. . . . .\(\displaystyle J\;=\;\sec\theta\tan\theta - \int\sec\theta(\sec^2\!\theta -1)\,d\theta\)

. . . . .\(\displaystyle J \;=\;\sec\theta\tan\theta - \int(\sec^3\!\theta - \sec\theta)\,d\theta\)

. . . . .\(\displaystyle J \;=\;\sec\theta\tan\theta - \underbrace{\int\sec^3\theta\,d\theta}_{\text{This is }J} + \int\sec\theta\,d\theta\)


\(\displaystyle \text{So we have: }\;J \;=\;\sec\theta\tan\theta - J + \ln|\sec\theta + \tan\theta| + C\)

. . . . . . . . \(\displaystyle 2J \;=\;\sec\theta\tan\theta + \ln|\sec\theta + \tan\theta| + C\)

. . . . . . . . .\(\displaystyle J \;=\;\tfrac{1}{2}\bigg[\sec\thgeta\tan\theta + \ln|\sec\theta + \tan\theta|\bigg] + C\)

\(\displaystyle \text{Therefore: }\;\boxed{\int\sec^3\!\theta\,d\theta \;=\;\tfrac{1}{2}\bigg[\sec\theta\tan\theta + \ln|\sec\theta + \tan\theta| \bigg] + C}\)



You have formulas for both integrals . . . integrate them, then back-substitute.

 
\(\displaystyle soroban, \ to \ take \ it \ one \ step \ further:\)

\(\displaystyle \int sec^{n}(\theta)d\theta \ = \ \int sec^{2}(\theta)sec^{n-2}(\theta)d\theta\)

\(\displaystyle I \ by \ P \ \implies \ \int \ udv \ = \ uv \ -\int vdu\)

\(\displaystyle Hence, \ let \ u \ = \ sec^{n-2}(\theta) \ \implies \ du \ = \ (n-2)sec^{n-3}(\theta)sec(\theta)tan(\theta)d\theta\)

\(\displaystyle = \ (n-2)sec^{n-2}(\theta)tan(\theta)d\theta, \ and \ dv \ = \ sec^{2}(\theta)d\theta \ \implies \ v \ = \ tan(\theta)\)

\(\displaystyle Therefore, \ \int sec^{n}(\theta)d\theta \ = \ sec^{n-2}(\theta)tan(\theta)-(n-2)\int sec^{n-2}(\theta)tan^{2}(\theta)d\theta\)

\(\displaystyle \int sec^{n}(\theta)d\theta \ = \ sec^{n-2}(\theta)tan(\theta)-(n-2)\int sec^{n-2}(\theta)[sec^{2}(\theta)-1]d\theta\)

\(\displaystyle \int sec^{n}(\theta)d\theta \ = \ sec^{n-2}(\theta)tan(\theta)-(n-2)\int [sec^{n}(\theta)-sec^{n-2}(\theta)]d\theta\)

\(\displaystyle \int sec^{n}(\theta)d\theta \ = \ sec^{n-2}(\theta)tan(\theta)-(n-2)\int sec^{n}(\theta)d\theta+(n-2)\int sec^{n-2}(\theta)d\theta\)

\(\displaystyle Ergo, \ (n-1)\int sec^{n}(\theta)d\theta \ = \ sec^{n-2}(\theta)tan(\theta)+(n-2)\int sec^{n-2}(\theta)d\theta\)

\(\displaystyle Finally, \ we \ have \ \int sec^{n}(\theta)d\theta \ = \ \frac{sec^{n-2}(\theta)tan(\theta)}{n-1}+\frac{(n-2)}{(n-1)}\int sec^{n-2}(\theta)d\theta, \ n \ \ne \ 1\)

\(\displaystyle Check: \ D_{\theta} \bigg[\frac{sec^{n-2}(\theta)tan(\theta)}{n-1}+\frac{n-2}{n-1}\int sec^{n-2}(\theta)d\theta\bigg] \ = \ sec^{n}(\theta)\)
 
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