# How to explain that $\int\frac{\sec\theta}{\tan^2\theta}d \theta$ is equal to $-\csc\theta+C$

While solving an integral doing trigonometric substitution, I ended up at solving

$$\int\frac{\sec\theta}{\tan^2\theta}d \theta$$

$$-\csc\theta+C$$

However, I could not figure out how to do it manually. I've first tried to transform the integral to $$\int\csc\theta\cot\theta$$ and then integrate by part, but the end result was not at all correct.

Now I suppose that the path I've chosen is incorrect, so please would someone point me in the right direction ?

Thanks.

• Integration by parts? You're not expected to know $\frac d{d\theta}(\csc\theta)=-\csc\theta\cot\theta$? – Mike Mar 14 '15 at 17:52
• @Mike if you read my comment at amWhy's answer, yes I already know that. However, I was actually trying to prove it, as MarioG did on his answer. – student Mar 14 '15 at 17:56
• When in my question I say that I know by head $\int\csc\theta\cot\theta = -\csc\theta + C$ that imply that I know by head the reverse too. – student Mar 14 '15 at 17:57
• I meant that it was considered a basic trigonometric derivative in Calc classes that I took, so I assumed it might in yours as well. Why reinvent the wheel? I mean shall we also prove that $\frac d{d\theta}\sin\theta=\cos\theta$? – Mike Mar 14 '15 at 18:05
• @Mike Why not ? It's not about re-inventing the wheel, it's just about having a deeper understanding. – student Mar 14 '15 at 18:32

\begin{align*} \int\dfrac{\sec \theta}{\tan^2 \theta}d\theta&=\int\dfrac{\frac{1}{\cos \theta}}{\frac{\sin^2\theta}{\cos^2\theta}}d\theta\\ &=\int(\sin \theta)^{-2}\cos \theta d\theta\\ &=-\frac{1}{\sin \theta} + C\\ &=-\csc \theta + C \end{align*}

• Exactly what I needed, will accept once I'm allowed to. – student Mar 14 '15 at 17:34

Rewrite the integral: $$\int\frac{\sec\theta}{\tan^2\theta}\mathrm d\mkern1.5mu \theta =\int\frac 1{\cos\theta\cfrac{\sin^2\theta}{\cos^2\theta}}\mathrm d\mkern1.5mu \theta=\int\frac{\cos\theta}{\sin^2\theta}\mathrm d\mkern1.5mu\theta$$ and set $u=\sin\theta$.

Hint:

Your integrand is by definition, by definition the derivative of $\csc\theta$ save for the sign. I.e.,

$$(\csc \theta)' = -\csc \theta\cot \theta$$

• You mean $-cscx$ ? And I know that, I'm more looking for how to prove it. – student Mar 14 '15 at 17:32
• Yes indeed! So $\int \csc \theta \cot \theta \,d\theta = - \csc \theta + C$ – Namaste Mar 14 '15 at 17:34