Trig Substitution

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goku900

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Mar 22, 2013
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The problem is

Integrate



To solve for this would I use a right triangle like so


I also think you might need to use a sec identity.

My friend gave me a hint and said x = (5 / 3 )sec(theta) but I do not know how he got that. If someone could explain that'd be nice.
 
Last edited:
goku900 said:
The problem is

Integrate sqrt(9x^2-25)/x^3 dx


To solve for this would I use a right triangle like so


I also think you might need to use a sec identity.

My friend gave me a hint and said x = (5 / 3 )sec(theta) but I do not know how he got that. If someone could explain that'd be nice.
Click to expand...

In general, when you want to deal with \(\displaystyle \sqrt{(a*x)^2 -b^2}\) you want to set

a*x = b * sec(Θ)

then you will get:

\(\displaystyle \sqrt{(a*x)^2 -b^2}\) = b * tan(Θ)
 
So does a = 3 because sqrt(9) = 3?

after I use what you said, I will end up plugging a tan(x) value for x?
 
Hello, goku900!

\(\displaystyle \displaystyle\text{Integrate: }\:\int \frac{\sqrt{9x^2-25}}{x^3}\,dx \)
Click to expand...
We want this triangle:
Code: 
[COLOR=#333333][SIZE=5][COLOR=#333333][SIZE=5]
                        *
                     *  *
             3x   *     *  ------
               *        * √9x²-25
            *           *[COLOR=#333333][SIZE=5]
         * θ            *
      *  *  *  *  *  *  *
                5[/SIZE][/COLOR][/SIZE][/COLOR][/SIZE][/COLOR]

We have: .\(\displaystyle \sec\theta \,=\,\frac{3x}{5} \quad\Rightarrow\quad x \,=\,\frac{5}{3}\sec\theta \quad\Rightarrow\quad dx \,=\,\frac{5}{3}\sec\theta\tan\theta\,d\theta\)

[COLOR=#de00e]. . \(\displaystyle \sqrt{9x^2-25} \:=\:\sqrt{25\sec^2\!\theta - 25} \:=\:\sqrt{25(\sec^2\!\theta - 1)} \:=\:\sqrt{25\tan^2\!\theta} \:=\:5\tan\theta \)


Substitute: .\(\displaystyle \displaystyle \int \frac{5\tan\theta}{\frac{125}{27}\sec^3\theta} \left(\tfrac{5}{3}\sec\theta\tan\theta\,d\theta \right) \;=\;\tfrac{9}{5}\int\frac{\sec\theta \tan^2\!\theta}{\sec^3\!\theta}\,d\theta \)

. . . . . . . . \(\displaystyle \displaystyle =\;\tfrac{9}{5}\int\frac{\tan^2\!\theta}{\sec^2\! \theta}\,d\theta \;=\; \tfrac{9}{5}\int \frac{\frac{\sin^2\!\theta}{\cos^2\!\theta}}{\frac{1}{\cos^2\!\theta}}\,d\theta \)

. . . . . . . . \(\displaystyle \displaystyle =\;\tfrac{9}{5}\int\sin^2\!\theta\,d\theta \;=\; \tfrac{9}{10}\int(1 - \cos2\theta)\,d\theta\)

Can you finish it now?[/COLOR]
 
soroban said:
Hello, goku900!

We have: .\(\displaystyle \sec\theta \,=\,\frac{3x}{5} \quad\Rightarrow\quad x \,=\,\frac{5}{3}\sec\theta \quad\Rightarrow\quad dx \,=\,\frac{5}{3}\sec\theta\tan\theta\,d\theta\)

[COLOR=#de00e]. . \(\displaystyle \sqrt{9x^2-25} \:=\:\sqrt{25\sec^2\!\theta - 25} \:=\:\sqrt{25(\sec^2\!\theta - 1)} \:=\:\sqrt{25\tan^2\!\theta} \:=\:5\tan\theta \)
[/COLOR]
Click to expand...

if x=5/3sec(theta)

how does the coefficient become 25. 9(5/3)=15..?
 
Hello, goku900!

If x = 5/3sec(theta), how does the coefficient become 25? . 9(5/3) = 15?
Click to expand...
You are not substituting correctly.

We have: .\(\displaystyle \sec\theta \,=\,\frac{3x}{5} \quad\Rightarrow\quad x \,=\,\frac{5}{3}\sec\theta \)

\(\displaystyle \sqrt{9x^2-25} \;=\;\sqrt{9\left(\frac{5}{3}\sec\theta\right)^2 - 25}\)

. . . . . . . . .\(\displaystyle =\;\sqrt{9\left(\color{red}{\frac{25}{9}}\sec^2\! \theta\right) - 25}\)

. . . . . . . . .\(\displaystyle =\;\sqrt{25\sec^2\!\theta - 25}\)

. . . . . . . . .\(\displaystyle =\;\sqrt{25(\sec^2\!\theta-1)}\)

. . . . . . . . .\(\displaystyle =\;\sqrt{25\tan^2\!\theta}\)

. . . . . . . . .\(\displaystyle =\;5\tan\theta\)
 
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