Intergration by Substition Formula compared to Symbolic Algebra Software Answer

• Jul 20th 2012, 04:42 PM
tammmyl
Intergration by Substition Formula compared to Symbolic Algebra Software Answer
For $\int{\sqrt(9-25x^2)} dx$

I got the answer by substitution into a formula:

$\int{\sqrt(9-25x^2)} dx = \frac{x}{2}\sqrt(9-25x^2)+\frac{9}{10}\arctan(\frac{5x}{\sqrt(9-25x^2)} + C$

By using Maple (symbolic algebra software) I got:-

$\int{\sqrt(9-25x^2)} dx = \frac{x}{2}\sqrt(9-25x^2)+\frac{9}{10}\arcsin(\frac{5}{3}x) + C$

Can anyone help me explain the difference in the two answers?
• Jul 20th 2012, 05:21 PM
skeeter
Re: Intergration by Substition Formula compared to Symbolic Algebra Software Answer
if $y = \arcsin\left(\frac{5x}{3} \right)$ ,

$\sin{y} = \frac{5x}{3} = \frac{opp}{hyp} \implies adj = \sqrt{9 - 25x^2}$

so,

$\tan{y} = \frac{opp}{adj} = \frac{5x}{\sqrt{9-25x^2}} \implies \arctan\left(\frac{5x}{\sqrt{9-25x^2}}\right) = y$

... in other words, they are two expressions for the same angle.
• Jul 20th 2012, 05:43 PM
tammmyl
Re: Intergration by Substition Formula compared to Symbolic Algebra Software Answer
Brilliant!! Thank you so much!
• Jul 20th 2012, 11:46 PM
Prove It
Re: Intergration by Substition Formula compared to Symbolic Algebra Software Answer
Quote:

Originally Posted by tammmyl
For $\int{\sqrt(9-25x^2)} dx$

I got the answer by substitution into a formula:

$\int{\sqrt(9-25x^2)} dx = \frac{x}{2}\sqrt(9-25x^2)+\frac{9}{10}\arctan(\frac{5x}{\sqrt(9-25x^2)} + C$

By using Maple (symbolic algebra software) I got:-

$\int{\sqrt(9-25x^2)} dx = \frac{x}{2}\sqrt(9-25x^2)+\frac{9}{10}\arcsin(\frac{5}{3}x) + C$

Can anyone help me explain the difference in the two answers?

When you have an integral of the form \displaystyle \begin{align*} f\left(a^2 - b^2x^2\right) \end{align*}, it's often useful to substitute \displaystyle \begin{align*} x = \frac{a}{b}\sin{\theta} \end{align*} or \displaystyle \begin{align*} x = \frac{a}{b}\cos{\theta} \end{align*}, because it simplifies nicely with the Pythagorean Identity. In this case, \displaystyle \begin{align*} \int{\sqrt{9 - 25x^2}\,dx} \end{align*}, we substitute \displaystyle \begin{align*} x = \frac{3}{5}\sin{\theta} \implies dx = \frac{3}{5}\cos{\theta}\,d\theta \end{align*} and the integral becomes

\displaystyle \begin{align*} \int{\sqrt{9 - 25x^2}\,dx} &= \int{\sqrt{9 - 25\left( \frac{3}{5}\sin{\theta} \right)^2}\,\frac{3}{5}\cos{\theta}\,d\theta} \\ &= \frac{3}{5}\int{\sqrt{9 - 9\sin^2{\theta}}\,\cos{\theta}\,d\theta} \\ &= \frac{3}{5}\int{ \sqrt{ 9\cos^2{\theta} } \, \cos{\theta} \, d\theta } \\ &= \frac{3}{5}\int{ 3\cos{\theta}\cos{\theta}\,d\theta } \\ &= \frac{9}{5}\int{\cos^2{\theta}\,d\theta} \\ &= \frac{9}{5}\int{\frac{1}{2} + \frac{1}{2}\cos{2\theta}\,d\theta} \\ &= \frac{9}{5}\left(\frac{1}{2}\theta + \frac{1}{4}\sin{2\theta}\right) + C \\ &= \frac{9}{10}\theta + \frac{9}{10}\sin{\theta}\cos{\theta} + C \\ &= \frac{9}{10}\theta + \frac{9}{10}\sin{\theta}\sqrt{1 - \sin^2{\theta}} +C \\ &= \frac{9}{10}\arcsin{\left(\frac{5}{3}x\right)} + \frac{9}{10}\left(\frac{5}{3}x\right)\sqrt{1 - \left(\frac{5}{3}x\right)^2} + C \\ &= \frac{9}{10}\arcsin{\left(\frac{5}{3}x\right)} + \frac{3}{2}x\sqrt{1 - \frac{25x^2}{9}} + C \end{align*}

\displaystyle \begin{align*} &= \frac{9}{10}\arcsin{\left(\frac{5}{3}x\right)} + \frac{3}{2}x\sqrt{\frac{9 - 25x^2}{9}} \\ &= \frac{9}{10}\arcsin{\left(\frac{5}{3}x\right)} + \frac{3}{2}x\,\frac{\sqrt{9 - 25x^2}}{3}\\ &= \frac{9}{10}\arcsin{\left(\frac{5}{3}x\right)} + \frac{1}{2}x\sqrt{9 - 25x^2} + C\end{align*}