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$$ \int \frac{3x^5}{\sqrt{2-x^2}} dx $$ Integrating the equation using integral-calculator gives me, $$ -\frac{\sqrt{2-x^2}(3x^4+8x^2+32)}{5} $$ But when I tried using trig identities I get this, $$ 3\int\frac{(\sqrt{2}sin\theta)^5}{\sqrt{2-2sin^2\theta}} \sqrt{2}cos\theta d\theta\\ 3\int\frac{(\sqrt{2}sin\theta)^5}{\sqrt{2(1-sin^2\theta)}} \sqrt{2}cos\theta d\theta\\ 3\int\frac{(\sqrt{2}sin\theta)^5}{\sqrt{2(cos^2\theta)}} \sqrt{2}cos\theta d\theta \\ 3*\sqrt{2}^5\int sin^5\theta d\theta\\ $$ Are they the same results? update: extended equation another update: i made up numbers...

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    $\begingroup$ The denominator is $\sqrt{2 - 2\sin^2\theta}$ and the integral becomes $3 (\sqrt{2})^5\int \sin^5 \theta d\theta$ $\endgroup$ – user348749 Feb 21 '17 at 8:25
  • $\begingroup$ What you did can't be right. You substitute $x=\sqrt2\sin\theta$ and then back with $\sin\theta=u$ (with $\sqrt2$ factored away), and end up with a simple power. This isn't possible. $\endgroup$ – Yves Daoust Feb 21 '17 at 8:32
  • $\begingroup$ @Muralidharan oic, thanks a lot $\endgroup$ – phoxd Feb 21 '17 at 8:33
  • $\begingroup$ @YvesDaoust now that I've edited, is the answer equivalent to using substitution right away? $\endgroup$ – phoxd Feb 21 '17 at 8:35
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    $\begingroup$ Incidentally, it is easier to do this without trigonometric substitution: If we set $u = 2 - x^2$, then $x^2 = 2 - u$ and $du = - 2x \,dx$, so the integral becomes $$\int \frac{3(x^2)^2 x \,dx}{\sqrt{2 - x^2}} = -\frac{3}{2} \int \frac{(2 - u)^2 \,du}{\sqrt{u}}.$$ Expanding the numerator lets us rewrite the integrand as a sum of power functions. $\endgroup$ – Travis Willse Feb 21 '17 at 11:00
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You must keep track of your substitutions that are used in helping you simplify the integral, so that you can return to your original variables.

Let $x = \sqrt{2} \sin \theta$. Then $dx = \sqrt{2} \cos \theta \,d\theta$. So

\begin{align*} \int \frac{3x^5}{\sqrt{2-x^2}} \,dx &= \int \frac{3(\sqrt{2} \sin \theta)^5}{\sqrt{2-2 \sin^2 \theta}} \cdot (\sqrt{2} \cos \theta) \,d\theta \\ &= \int \frac{12\sqrt{2} \sin^5 \theta}{\sqrt{2 \cos^2 \theta}} \cdot (\sqrt{2} \cos \theta) \,d\theta \\ &= \int \frac{12\sqrt{2} \sin^5 \theta}{\sqrt{2} \cos \theta} \cdot (\sqrt{2} \cos \theta) \,d\theta \\ &= 12\sqrt{2} \int \sin^5 \theta \,d\theta \\ &= 12\sqrt{2} \int (\sin^2 \theta)^2 \sin \theta \,d\theta \\ &= 12\sqrt{2} \int (1 - \cos^2 \theta)^2 \sin \theta \,d\theta \end{align*}

Let $u = \cos \theta$. Then $du = - \sin \theta \, d\theta$.

\begin{align*} \int \frac{3x^5}{\sqrt{2-x^2}} \,dx &= 12\sqrt{2} \int (1 - \cos^2 \theta)^2 \sin \theta \,d\theta \\ &= -12\sqrt{2} \int (1 - u^2)^2 \,du \\ &= -12\sqrt{2} \int (1 - 2u^2 + u^4)\,du \\ &= -12\sqrt{2} \left( u - \frac{2u^3}{3} + \frac{u^5}{5} \right) + C \\ &= -12\sqrt{2} \left( \cos \theta - \frac{2}{3} \cos^3 \theta + \frac{1}{5} \cos^5 \theta \right) + C \end{align*}

Recall $x = \sqrt{2} \sin \theta$. Consider a right triangle with angle $\theta$ touching the hypotenuse, side opposite this angle length $x$, and hypotenuse length $\sqrt{2}$. Then we see that $$\sin \theta = \frac{x}{\sqrt{2}} \qquad \cos \theta = \frac{\sqrt{2-x^2}}{\sqrt{2}} $$ So \begin{align} \int \frac{3x^5}{\sqrt{2-x^2}} \,dx &= -12\sqrt{2} \left( \cos \theta - \frac{2}{3} \cos^3 \theta + \frac{1}{5} \cos^5 \theta \right) + C \\ &= -12\sqrt{2} \left( \frac{(2-x^2)^{1/2}}{\sqrt{2}} - \frac{2}{3} \frac{(2-x^2)^{3/2}}{2\sqrt{2}} + \frac{1}{5}\frac{(2-x^2)^{5/2}}{4\sqrt{2}} \right) + C \\ &= -12 \left( (2-x^2)^{1/2} - \frac{1}{3} (2-x^2)^{3/2} + \frac{1}{20}(2-x^2)^{5/2} \right) + C \\ &= -12 \sqrt{2-x^2} \left( 1 - \frac{1}{3} (2-x^2) + \frac{1}{20}(2-x^2)^{2} \right) + C \\ &= \sqrt{2-x^2} \left(-12 + 4(2-x^2) - \frac{3}{5}(2-x^2)^{2} \right) + C \\ &= \frac{\sqrt{2-x^2}}{5} \left(-60 + 20(2-x^2) - 3(2-x^2)^{2} \right) + C \\ &= \frac{\sqrt{2-x^2}}{5} \left(-60 + 40 - 20x^2 - 12 + 12x^2 - 3x^4 \right) + C \\ &= \frac{\sqrt{2-x^2}}{5} \left(-32 - 8x^2 - 3x^4 \right) + C \\ &= -\frac{\sqrt{2-x^2}}{5} \left(3x^4 + 8x^2 + 32 \right) + C \end{align}

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  • $\begingroup$ Thanks a lot for a detailed description! $\endgroup$ – phoxd Feb 21 '17 at 9:16
  • $\begingroup$ @PhoxKiD No problem! $\endgroup$ – Manuel Guillen Feb 21 '17 at 9:29
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\begin{align*} \int \sin^5 \theta d\theta &= -\int (1-\cos^2\theta)^2 (-\sin \theta) d\theta\\ &=-\int 1-2u^2 + u^4 du, u = \cos \theta\\ &= - \left(u - \frac{2u^3}{3} + \frac{u^5}{5}\right)\\ &= -\cos\theta\left(\frac{15 - 10 \cos^2 \theta + 3\cos^4\theta}{15}\right)\\ &= -\sqrt{1-\sin^2\theta} \left(\frac{8 + 4\sin^2\theta + 3\sin^4\theta}{15}\right) \end{align*} Now substitute $\sin \theta = x/\sqrt{2}$

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$$\int\sin^5t\,dt=\int(1-\cos^2t)^2\sin t\,dt=-\int(1-z^2)^2\,dz=-\int(1-2z^2+z^4)^2\,dz\\ =-z+\frac23z^3-\frac{z^5}5=-\cos t+\frac23\cos^3t-\frac{\cos^5t}5\\ =-\sqrt{1-\sin^2t}\left(1-\frac23(1-\sin^2t)+\frac{(1-\sin^2t)^2}5\right)\\ =-\sqrt{1-\frac{x^2}2}\left(1-\frac23\left(1-\frac{x^2}2\right)+\frac{\left(1-\dfrac{x^2}2\right)^2}5\right).$$

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