How to simplify $\sqrt{\tan^2\theta}\sin\theta\cos^2\theta$? 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.

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.
