Question

Integration \(\int\limits_0^{2\pi } {si{n^2}\left( {\frac{\theta }{3}} \right)} d\theta \).

Short answer

\(\pi + \frac{{3\sqrt 3 }}{8}\) is the final answer to the question.

Step-by-step solution

Formula Used.

\({\sin ^2}\theta = \frac{{1 - \cos \,2\theta }}{2}\)

\({\sin ^2}\left( {\frac{\theta }{3}} \right) = \left( {\frac{{1 - \cos \frac{{2\theta }}{3}}}{2}} \right)\)

Calculate Integral.

\( = \int\limits_0^{2\pi } {\left( {\frac{{1 - \cos \frac{{2\theta }}{3}}}{3}} \right)} d\theta \)

\( = \frac{1}{2}\int\limits_0^{2\pi } {\left( {1 - \cos \frac{{2\theta }}{3}} \right)} d\theta \) \(\begin{aligned}{l}\int {d\theta = \theta } \\\int {\cos \,\frac{{2\theta }}{3} = \sin \frac{{2\theta }}{3} \times \frac{3}{2}} \end{aligned}\)

\( = \frac{1}{2}\left( {\theta - \frac{3}{2}\sin \frac{{2\theta }}{3}} \right)_0^{2\pi }\)

Now put the Upper limit-Lower limit.

\( = \frac{1}{2}\left( {2\pi - 0 - \frac{3}{2}\sin \frac{{4\pi }}{3} - 0} \right)\)

\(\begin{aligned}{l} &= \frac{1}{2}\left( {2\pi - \frac{3}{2} \times \left( { - \frac{{\sqrt 3 }}{2}} \right)} \right)\\ &= \left( {\pi + \frac{{\sqrt 3 \times 3}}{8}} \right)\end{aligned}\)

Hence, \( = \pi + \frac{{\sqrt 3 \times 3}}{8}\) this is the final answer.