Question

Evaluate the integral if exists.

\(\int_{\rm{0}}^{\rm{1}} {{\rm{sin}}} {\rm{(3\pi t)dt}}\)

Short answer

The required solution of the integral is\(\int_0^1 {\sin } (3\pi t)dt = \frac{2}{{3\pi }}\).

Step-by-step solution

Methods of evaluating integral

For evaluating integrals there are various methods like substitution (also called u-substitution), integration by parts, partial fractions etc.

Applying substitution method and setting new limits

Let \(u = 3\pi t\)

\(\begin{aligned}{c}du &= 3\pi {\rm{ }}dt\\dt &= \frac{1}{{3\pi }}du\end{aligned}\)

Now new limit of integration for u is

\(u = \left\{ {\begin{array}{*{20}{l}}{3\pi }&{,{\rm{ }}if{\rm{ }}t = 1}\\0&{,{\rm{ }}if{\rm{ }}t = 0}\end{array}} \right.\)

Substituting u in given function to get required results

Substitute \(u = 3\pi t{\rm{ and }}dt = \frac{1}{{3\pi }}du\)

Therefore, \(\int_0^1 {\sin } (3\pi t)dt = \frac{2}{{3\pi }}\).