# Thread: integration by trig substitution giving me different, equivalent answers, as calcula?

1. ## integration by trig substitution giving me different, equivalent answers, as calcula?

Hi. Using trig sub to get integrals, I keep getting different but possibly equivalent answers as in book, and on calculator.

For example, integral(5/sqrt(25x^2-9),x)

By trig sub,

I get: ln|5x/3+(5sqrt(x^2-9/25)/3|+c

Calculator gets: ln(sqrt(25x^2-9)+5x)

Wolfram integrator gets: log(2(sqrt(25x^2-9)+5x))

But, I am pretty sure I did the problem correctly.

Are they equivalent? If they are, How can I show that they are to avoid my teacher marking off points?

2. ## Re: integration by trig substitution giving me different, equivalent answers, as calc

Hint :

Substitute $\displaystyle ~x=\frac{3}{5\cdot \cos u}$

Result should be :

$\displaystyle I= \ln (\sqrt{25x^2-9}+5x) + C$

3. ## Re: integration by trig substitution giving me different, equivalent answers, as calc

Originally Posted by princeps
Hint :

Substitute $\displaystyle ~x=\frac{3}{5\cdot \cos u}$

Result should be :

$\displaystyle I= \ln (\sqrt{25x^2-9}+5x) + C$
Thats what I did.

The integral simplifies first to integral(1/sqrt(x^2-9/25),x)

x=(3/5)*sec(u)
...

In the end it is integral(sec(u),u)

=ln|sec(u)+tan(u)|+c

I have a right triangle, where hypotenuse is x, opposite is sqrt(x^2-9/25), and adjacent=3/5.

So, sec(u)=1/((3/5)/x)=5x/3, and tan(u)=sqrt(x^2-9/25)/(3/5)=sqrt(25x^2-9)/3.

So I get: $\displaystyle I= \ln (\sqrt{25x^2-9}/3+5x/3)+ C$

4. ## Re: integration by trig substitution giving me different, equivalent answers, as calc

Originally Posted by StudentMCCS
Thats what I did.

The integral simplifies first to integral(1/sqrt(x^2-9/25),x)

x=(3/5)*sec(u)
...

In the end it is integral(sec(u),u)

=ln|sec(u)+tan(u)|+c

I have a right triangle, where hypotenuse is x, opposite is sqrt(x^2-9/25), and adjacent=3/5.

So, sec(u)=1/((3/5)/x)=5x/3, and tan(u)=sqrt(x^2-9/25)/(3/5)=(5*sqrt(25x^2-9))/3.

So I get: $\displaystyle I= \ln (\sqrt{25x^2-9}/3+5x/3)+ C$
You did everything ok .

$\displaystyle I= \ln (\sqrt{25x^2-9}/3+5x/3)+ C_1=\ln (\sqrt{25x^2-9}+5x)+ \ln \frac{1}{3}+ C_1 ~$ , hence :

$\displaystyle I= \ln (\sqrt{25x^2-9}+5x)+ C$

5. ## Re: integration by trig substitution giving me different, equivalent answers, as calc

Originally Posted by princeps
You did everything ok .

$\displaystyle I= \ln (\sqrt{25x^2-9}/3+5x/3)+ C_1=\ln (\sqrt{25x^2-9}+5x)+ \ln \frac{1}{3}+ C_1 ~$ , hence :

$\displaystyle I= \ln (\sqrt{25x^2-9}+5x)+ C$
Ahh, that makes sense. Thanks. I wonder which way my teacher would prefer I wrote it?