Question

Evaluate the indefinite integral\(\int {\sin \pi tdt} \).

Short answer

The indefinite integral value of the given equation is\(\int {(\sin \pi t)dt} = - \frac{1}{\pi }\cos \pi t + c\).

Step-by-step solution

Definition of the indefinite integral

An integral that has no upper bound and no lower bound is known as indefinite integral.

Expand the number of variables in the equation.

Given:

\(I = \int {(\sin \pi t)dt} \) …(1)

Let \(u = \pi t\)

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

Substitute \(\pi t\)for u and \(dt = \frac{{du}}{\pi }\) for \({\rm{dt}}\)in the equation (1).

\(I = \frac{1}{\pi }\int {(\sin u)du} \)

Evaluate the equation.

Integrate

\(I = - \frac{1}{\pi }\cos u + c\)

Return to the original position\(u = \pi t\)

\(I = - \frac{1}{\pi }\cos \pi t + c\)

Therefore, the indefinite integral value of the given equation is\(\int {(\sin \pi t)dt} = - \frac{1}{\pi }\cos \pi t + c\).