Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a complicated integral to solve. Can someone provide a better way to solve it than what i did - dividing by a inside the root, and then putting $ t = x / a $, and then putting $t^2 = \cos \theta $ and so many other substitutions. $$ \int x \sqrt{\frac {a^2 - x^2} { a^2 + x^2 }} dx $$

share|improve this question

2 Answers 2

Let $x^2=a^2\cos(y)$. We then have $2xdx = -a^2 \sin(y) dy$. The integral then becomes \begin{align} I & = \int x \sqrt{\dfrac{a^2-x^2}{a^2+x^2}} dx = -\int \dfrac{a^2}2 \sin(y) \sqrt{\dfrac{1-\cos(y)}{1+\cos(y)}}dy = -\dfrac{a^2}2 \int \sin(y) \tan(y/2) dy\\ & = -a^2\int \sin^2(y/2)dy \end{align} I trust you can finish it from here.

share|improve this answer
    
Thanks! That's better :D –  Sri Krishna Apr 24 '13 at 7:47
    
@SriKrishna You might be slightly bothered by the different signs for the integral, but the limits will take care of the signs. So be careful with the limits. –  user17762 Apr 24 '13 at 7:49

We can use the algebraic substitution:

$$t=\frac{a^{2}-x^{2}}{a^{2}+x^{2}}.\tag{0} $$

We have that

$$\begin{eqnarray*} I &=&\int x\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}dx=-\int \sqrt{t}\frac{a^{2} }{\left( 1+t\right) ^{2}}dt\tag{1} \\ &=&a^{2}\frac{\sqrt{t}}{1+t}-a^{2}\arctan \sqrt{t}+C \\ &=&\frac{a^{2}+x^{2}}{2}\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}-a^{2}\arctan \sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}+C, \end{eqnarray*}$$

because the integral in $t$ $(1)$ can be evaluated by using another algebraic substitution, $$u^{2}=t,\tag{2}$$ and expanding into partial fractions the resulting integrand $$ \begin{eqnarray*} \int \frac{\sqrt{t}}{\left( 1+t\right) ^{2}}dt &=&2\int \frac{u^{2}}{\left( 1+u^{2}\right) ^{2}}\,du \\ &=&2\int -\frac{1}{\left( 1+u^{2}\right) ^{2}}+\frac{1}{1+u^{2}}\,du.\tag{3} \end{eqnarray*} $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.