# How to evaluate the limits of $\int (\sin y) dy$?

How to evaluate this integral of $$\int^{\arctan2}_{\arctan \frac{1}{2}} \sin y ~dy$$?

Now I know this is $$-[\cos(\arctan 2)-\cos(\arctan 0.5)]=[\cos(\arctan 0.5)-\cos(\arctan 2)]$$ but I have no idea how to figure this out without a calculator.

• Always trace out triangles and solve whenever inverse trigonometry values are involved. – Sujith Sizon Oct 11 '15 at 2:52

## 2 Answers

Look at a $1,2,\sqrt{5}$ triangle and all will reveal itself.

So in my answer I mention the significance of triangles. Another poster has so I'll reiterate.

For $\arctan(\frac{1}{2})$ draw a right angled triangle, denote one side (not the hyp) to be of length 2, and the other not-hyp as length 1.

Which angle is $\arctan(\frac{1}{2})$ - good, now what's the $\cos$ of this angle.

Repeat or relate $\tan(\theta)$ to $\tan(\frac{1}{\theta})$ to complete answer.

(SE is against doing homework questions)