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

$$ \int\frac {11 \tan^2\theta}{\sqrt{49 - \tan^2\theta}} \Big( \sec^2\theta\,d\theta\Big) \text{ (So } \sec^2\theta\,d\theta \text{ becomes } du.) $$ — Michael Hardy, Jul 22 '17 at 2:14
Type (11sec²(x)tan²(x))/(49-tan²(x))^(1/2) into www.integral-calculator.com for a FULL explanation after you click “show steps” — gen-ℤ ready to perish, Jul 22 '17 at 5:37

APCorreia · Accepted Answer · 2017-07-22 15:50:23Z

11

HINT

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

answered Jul 22 '17 at 2:11

APCorreia

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2

$\begingroup$ that $49$ was a dead giveaway, wasn't it? $\endgroup$ – AccidentalFourierTransform Jul 22 '17 at 9:42

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Harry Peter · Accepted Answer · 2017-07-24 17:06:35Z

$\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) $

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

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

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

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

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

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

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

2 Answers 2