calculus - Evaluate the $int_0^{1}{cos(frac{pi t}2)}dt$

Evaluate the definite integral

$$\int_0^{1}{\cos(\frac{\pi t}2)}dt$$

I've been indefinite intervals like this:

$$\int{\frac{\cos x}{\sin ^2x}}dt$$ so I could do this:

$$u=sinx$$

$$du=cosx ....$$

And things would workout, but with: $$\int_0^{1}{\cos(\frac{\pi t}2)}dt$$ I'm having troubles figuring out what to substitute $$u=\frac{\pi t}{2}$$ Doesn't seem right because then

$$du=\frac\pi 2$$ And that doesn't fit in my integral anywhere.

Is this right?

So $$\frac{\sin u}{du}$$

$$=\frac{\sin \frac {\pi t} 2}{\frac \pi 2} | f(1) - f(0)$$

$$\frac{\sin \frac {\pi (1)} 2}{\frac \pi 2} - 0$$

$$= \frac 2 \pi$$

Answer

Try using subsitution rule.

$$u = \frac{\pi}2 t \text{ and } du = \frac{\pi}2 \, dt \implies \frac 2{\pi} du = dt$$

And since this is a definite integral, change your limits accordingly: $$u(0)=\frac{\pi}2 \cdot 0=0 \text{ and }u(1)=\frac{\pi}2 \cdot 1=\frac{\pi}2$$

Finally, $$\begin{align*} \int_0^1 \cos\left(\frac {\pi}2 t \right) \, dt&=\frac 2{\pi} \int_0^{\pi/2} \cos u \, du \\ \end{align*}$$ Can you take it from here?