A very tricky integral

Question

I am trying to integrate

$$ \int \frac{11\sec^2\theta\tan^2\theta}{\sqrt{49-\tan^2\theta}}d\theta $$

and I am continually getting the wrong answer. I would like steps in solving this problem

Answer 1

HINT

Make the substitution $\displaystyle u = \frac{\tan(\theta)}{7}$.

Answer 2

$\int\dfrac{11\sec^2\theta\tan^2\theta}{\sqrt{49-\tan^2\theta}}~d\theta $

$=\int\dfrac{11\tan^2\theta}{\sqrt{49-\tan^2\theta}}~d(\tan\theta) $

$=\int\dfrac{11x^2}{\sqrt{49-x^2}}~dx$ $(\text{Let}~x=\tan\theta)$

$=\int77\sin^2\phi~d\phi$ $(\text{Let}~x=7\sin\phi)$

$=\int\dfrac{77-77\cos2\phi}{2}~d\phi$

$=\dfrac{77\phi}{2}-\dfrac{77\sin2\phi}{4}+C$

$=\dfrac{77\phi}{2}-\dfrac{77\sin\phi\cos\phi}{2}+C$

$=\dfrac{77}{2}\sin^{-1}\dfrac{x}{7}-\dfrac{11x\sqrt{49-x^2}}{14}+C$

$=\dfrac{77}{2}\sin^{-1}\dfrac{\tan\theta}{7}-\dfrac{11\sqrt{49-\tan^2\theta}\tan\theta}{14}+C$