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To begin with the $d\theta$ on the top of the fraction threw me off but I'm assuming it's just another way of representing:

$${\int}\frac{1}{9\cos^2\theta+\sin^2\theta}\,d\theta$$

I tried working backwards

$$\frac{d}{d\theta}\tan\theta=\sec^2\theta\,\,\,\,{\Rightarrow}\,\,\,\,d\,\tan\theta=\sec^2\theta\,d\theta$$ $${\Rightarrow}\,{\int}\frac{\sec^2\theta\,d\theta}{9+\tan^2\theta}$$

$$\tan\theta=\frac{\sin\theta}{\cos\theta}\,\,\,\,{\Rightarrow}\,\,\,\,\tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}$$

$${\Rightarrow}\,{\int}\frac{\sec^2\theta\,d\theta}{\left(9+\dfrac{\sin^2\theta}{\cos^2\theta}\right)}$$

$$9=\frac{9\cos^2\theta}{\cos^2\theta}\,\,\,\,{\Rightarrow}\,\,\,\,{\int}\frac{\sec^2\theta\,d\theta}{\left(\dfrac{9\cos^2\theta}{\cos^2\theta}+\dfrac{\sin^2\theta}{\cos^2\theta}\right)}\,\,\,\,{\Rightarrow}\,\,\,\,{\int}\frac{\sec^2\theta\,d\theta}{\left(\dfrac{9\cos^2\theta+\sin^2\theta}{\cos^2\theta}\right)}$$

$${\Rightarrow}\,\,\,\,{\int}\frac{\color{red}{\cos^2\theta\,\sec^2\theta}\,d\theta}{9\cos^2\theta+\sin^2\theta}$$

Now I have to prove that $$\cos^2\theta\,\sec^2\theta=1$$ but I don't think it is... What have I done wrong? Regards Tom

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    $\begingroup$ Define $\cos\theta,\sec\theta$ $\endgroup$ Commented Apr 8, 2015 at 11:49
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    $\begingroup$ $sec^2\theta=\frac{1}{cos^2\theta} \implies cos^2\theta sec^2\theta = 1$ $\endgroup$
    – Waffle
    Commented Apr 8, 2015 at 11:56
  • $\begingroup$ @Waffle Could you please convert your comment into an answer so this question can be removed from the "Unanswered" queue? $\endgroup$ Commented Nov 15, 2018 at 5:14
  • $\begingroup$ @Waffle Thanks! $\endgroup$ Commented Nov 16, 2018 at 1:00

2 Answers 2

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$\sec^2\theta = \frac{1}{\cos^2\theta} \implies \cos^2\theta\sec^2\theta = \frac{\cos^2\theta}{\cos^2\theta} = 1$

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An alternative method. Let $$I={\int}\frac{1}{9\cos^2x+\sin^2x}\,dx$$ Dividing the denominator and numerator by $\cos^2x$ we have $$I=\int\frac{\sec^2x}{9+\tan^2x}dx$$ Make the substitution $t=\tan x$, so that $\sec^2x~dx=dt,$ and we are done.

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