Question

\({\bf{1 - 40}}\)- Evaluate the integral.

\(\int_0^{\frac{\pi }{6}} t \sin 2tdt\)

Short answer

\(\int_0^{\frac{\pi }{6}} t \sin 2tdt = \frac{{3\sqrt 3 - \pi }}{{24}}\)

Step-by-step solution

Definition

Integrationis a method of adding or summing up the parts to find the whole. It is a reverse process of differentiation.

Evaluate integral

Consider the definite integral\(\int_0^{\frac{\pi }{6}} t \sin 2tdt\)

Using integration by parts\(\int v dv = uv - \int v du\,\, - - - - (1)\)

\(\int_0^{\frac{\pi }{6}} t \sin 2tdt = \left( { - \frac{1}{2}t\cos 2t - \int - \frac{1}{2}\cos 2tdt} \right)_0^{\frac{\pi }{6}}\)

Applying limits

Applying lower & upper limit we have:

\(\begin{aligned}{c}\left( { - \frac{1}{2}t\cos 2t + \frac{1}{4}\sin 2t} \right)_0^{\frac{\pi }{6}} &= \left( { - \frac{1}{2}\left( {\frac{\pi }{6}} \right)\left( {\cos \left( {\frac{\pi }{3}} \right)} \right) + \frac{1}{4}\left( {\sin \frac{\pi }{3}} \right) - \left( { - \frac{1}{2}(0)\cos 0 + \frac{1}{4}\sin 0} \right)} \right)\\ &= - \frac{\pi }{{12}}\left( {\frac{1}{2}} \right) + \frac{1}{4}\left( {\frac{{\sqrt 3 }}{2}} \right) + 0 - \frac{1}{4}(0)\\ = \frac{{ - \pi }}{{24}} + \frac{{\sqrt 3 }}{8}\\ &= \frac{{3\sqrt 3 - \pi }}{{24}}\end{aligned}\)

Therefore, the value of the given integral is\(\frac{{3\sqrt 3 - \pi }}{{24}}\).