# Going from integral of $dx$ to integral of $d\theta$?

If I'm substituting $$x = \tan(\theta) \qquad \text{into} \qquad \int \frac{1}{(1+x^2)^2} \, dx$$

I get $$\int \cos^4(\theta) \, dx \quad \text{but the correct answer is} \quad \int \cos^2(\theta) \, d\theta$$

I don't understand why because $$(1 + \tan^{2}(\theta)) = \sec^2(\theta) \qquad\text{so}\qquad \frac{1}{(\sec^2(\theta))^2} \quad \text{should be} \quad \cos^4(\theta)$$

right?

I don't understand what I'm doing wrong.

• You forgot about $dx$ – openspace Apr 10 '16 at 18:57
• $dx$ = $cos^{2}{t} dt$ – openspace Apr 10 '16 at 18:58
• When you write "but the correct answer is" inside the MathJax code you can write \text{ but the correct answer is }. I changed it to that. That way you don't need to manually add spaces between words. \text{} is the standard control sequence used for that purpose. $\qquad$ – Michael Hardy Apr 10 '16 at 19:21
• Thanks! That's very useful. Sorry about that, this is my first time using MathJax. – user2513924 Apr 10 '16 at 20:05

You forgot about the $dx$, it would be $\sec^2 \theta \, d \theta$, So doing the required cancellations, you would get $$\int \cos^2 \theta \, d\theta$$ This can be tackled by writing $\cos^2 \theta$ as $\dfrac{1+\cos 2\theta }{2}$ and then integrating.

• Thanks! I hadn't realised that you would have to do anything about the d-anything. – user2513924 Apr 10 '16 at 20:11
• @user2513924 You always have to do something about it! – Nikunj Apr 10 '16 at 20:45