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I am solving $\int_{7}^{14}\frac{\sqrt{x^2-49}}{x^4}$ and got the integral down to $\frac{1}{343}\int_{0}^{\frac{\pi}{3}}\sqrt{\tan^2\theta}\sin\theta\cos^2\theta$ and wolfram simplified $\sqrt{\tan^2\theta}\sin\theta\cos^2\theta$ to $\sin^2(\theta)\cos(\theta)$ and I cannot figure out how to do this, thanks for all the help in advance.

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$\sqrt{\tan^{2}\theta}=\left|\tan\theta\right|$ and on interval $\left[0,\frac{\pi}{3}\right]$ we have $\left|\tan\theta\right|=\tan\theta=\frac{\sin\theta}{\cos\theta}$ leading to $$\sqrt{\tan^{2}\theta}\sin\theta\cos^{2}\theta=\sin^{2}\theta\cos\theta$$

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  • $\begingroup$ +1 for mentioning the fact that $\sqrt{\tan^2\theta}=|\tan\theta|=\tan\theta$ on $[0,\pi/3].$ $\endgroup$ – Gahawar Sep 7 '14 at 19:32
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Note that $\tan^2 \theta = (\tan\theta)^2$, and likewise for $\cos^2 \theta$, and on the interval of integration, we have that $\tan\theta \geq 0$.

$$\sqrt{\tan^2 \theta}\sin \theta \cos^2 \theta = \tan\theta\cdot \sin \theta \cos^2 \theta$$

$$= \frac{\sin\theta}{\cos\theta}\cdot \sin\theta \cos^2 \theta $$

$$ = \sin^2 \theta \cos\theta$$

Now put $u = \sin\theta \implies du = \cos \theta d\theta$ and use the power-rule to integrate.

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    $\begingroup$ Noting carefully that $\sin \theta$ and $\cos \theta$ are non-negative between the limits of integration. $\endgroup$ – Mark Bennet Sep 7 '14 at 19:28
  • $\begingroup$ Ahhh I got confused since wolfram simplified it to$\sqrt{\tan^2\theta}\sin\theta(1-\sin^2\theta)$ and then got their answer and I was trying to solve from that. but there wasn't any need to use the pythagorean identity. Thanks for the help! $\endgroup$ – Kenshin Sep 7 '14 at 19:30

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