Calculus 2 integrals.

• Feb 12th 2013, 11:24 PM
danielhigareda
Calculus 2 integrals.
need help i can not figure this question out. I know I got the wrong answer just dont know what I did wrong.

The question is integral of (x^3)/((3+x^2)^(5/2)) from 0 to 3. using the right triangle trig
• Feb 12th 2013, 11:46 PM
danielhigareda
Re: Calculus 2 integrals.
• Feb 13th 2013, 12:27 AM
JJacquelin
Re: Calculus 2 integrals.
• Feb 13th 2013, 12:30 AM
Prove It
Re: Calculus 2 integrals.
Quote:

Originally Posted by danielhigareda
need help i can not figure this question out. I know I got the wrong answer just dont know what I did wrong.

The question is integral of (x^3)/((3+x^2)^(5/2)) from 0 to 3. using the right triangle trig

Let \displaystyle \begin{align*} x = \sqrt{3} \, \tan{(\theta)} \implies dx = \sqrt{3}\, \sec^2{(\theta)}\,d\theta \end{align*} note that when \displaystyle \begin{align*} x = 0, \theta = 0 \end{align*} and when \displaystyle \begin{align*} x = 3, \theta = \frac{\pi}{3} \end{align*}, then the integral becomes

\displaystyle \begin{align*} \int_0^3{\frac{x^3}{\left( 3 + x^2 \right)^{\frac{5}{2}}}\,dx} &= \int_0^{\frac{\pi}{3}}{\frac{\left[ \sqrt{3}\,\tan{(\theta)} \right] ^3}{\left[ 3 + \left( \sqrt{3} \, \tan{(\theta)} \right)^2 \right]^{ \frac{5}{2} } }\,\sqrt{3}\,\sec^2{(\theta)}\,d\theta} \\ &= \int_0^{\frac{\pi}{3}}{\frac{9\tan^3{(\theta)}\sec ^2{(\theta)}}{ \left[ 3 + 3\tan^2{(\theta)} \right] ^{ \frac{5}{2} } }\,d\theta} \\ &= \int_0^{\frac{\pi}{3}}{\frac{9\tan^3{(\theta)}\sec ^2{(\theta)}}{9\sqrt{3} \left[ 1 + \tan^2{(\theta)} \right] ^{\frac{5}{2}} }\,d\theta} \\ &= \frac{1}{\sqrt{3}} \int_0^{\frac{\pi}{3}}{\frac{\tan^3{(\theta)}\sec^ 2{(\theta)}}{\left[ \sec^2{(\theta)} \right] ^{\frac{5}{2}} }} \\ &= \frac{1}{\sqrt{3}} \int_0^{\frac{\pi}{3}}{\frac{\tan^3{(\theta)}\sec^ 2{(\theta)}}{ \sec^5{(\theta)} }\,d\theta} \\ &= \frac{1}{\sqrt{3}} \int_0^{\frac{\pi}{3}}{ \frac{ \tan^3{(\theta)} }{ \sec^3{(\theta)} } \, d\theta } \end{align*}

\displaystyle \begin{align*} &= \frac{1}{\sqrt{3}} \int_0^{\frac{\pi}{3}}{\sin^3{(\theta)}\,d\theta} \\ &= -\frac{1}{\sqrt{3}} \int_0^{\frac{\pi}{3}}{-\sin{(\theta)} \left[ 1 - \cos^2{(\theta)} \right] d\theta } \end{align*}

Now let \displaystyle \begin{align*} u = \cos{(\theta)} \implies du = -\sin{(\theta)}\,d\theta \end{align*} and note that \displaystyle \begin{align*} u(0) = 1 \end{align*} and \displaystyle \begin{align*} u \left( \frac{\pi}{3} \right) = \frac{1}{2} \end{align*} and the integral becomes

\displaystyle \begin{align*} -\frac{1}{\sqrt{3}} \int_0^{\frac{\pi}{3}}{-\sin{(\theta)} \left[ 1 - \cos^2{(\theta)} \right] d\theta } &= -\frac{1}{\sqrt{3}} \int_1^{\frac{1}{2}}{ 1 - u^2 \, du } \\ &= \frac{1}{\sqrt{3}} \int_{\frac{1}{2}}^1{ 1 - u^2 \, du } \\ &= \frac{1}{\sqrt{3}} \left[ u - \frac{u^3}{3} \right]_{\frac{1}{2}}^1 \\ &= \frac{1}{\sqrt{3}} \left\{ \left[ 1 - \frac{1^3}{3} \right] - \left[ \frac{1}{2} - \frac{ \left( \frac{1}{2} \right)^3 }{3} \right] \right\} \\ &= \frac{1}{\sqrt{3}} \left[ \left( 1 - \frac{1}{3} \right) - \left( \frac{1}{2} - \frac{1}{24} \right) \right] \\ &= \frac{1}{\sqrt{3}} \left( \frac{2}{3} - \frac{11}{24} \right) \\ &= \frac{1}{\sqrt{3}} \left( \frac{5}{24} \right) \\ &= \frac{5}{24 \sqrt{3} } \\ &= \frac{5\sqrt{3}}{72} \end{align*}
• Feb 15th 2013, 10:34 PM
RolandoGamble
Re: Calculus 2 integrals.