Definite Integration Question - None

Question

If \int_0^{\pi/3} \frac{ an heta}{2k\sec heta} d heta = 1 - \frac{1}{\sqrt{2}}, (k > 0), then the value of k is

Accepted Answer

Simplify: \frac{ an heta}{2k\sec heta} = \frac{\sin heta}{2k\cos heta} \cdot \cos heta = \frac{\sin heta}{2k}

So: \int_0^{\pi/3} \frac{\sin heta}{2k} d heta = \frac{1}{2k}[-\cos heta]_0^{\pi/3} = \frac{1}{2k}\left(-\frac{1}{2} + 1 ight) = \frac{1}{4k}

Setting equal: \frac{1}{4k} = 1 - \frac{1}{\sqrt{2}}, solve for k.

Question: