0
$\begingroup$

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

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

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

$$= \frac 2 \pi$$

$\endgroup$
4
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

HINT:

For $a\ne0,$ $$\int \cos at\ dt=\frac{\sin at}a+K$$

$\endgroup$
  • $\begingroup$ Wish I know the mistake here? $\endgroup$ – lab bhattacharjee Aug 5 '14 at 8:49
1
$\begingroup$

Given

$$\int_0^1\cos\left( \frac{\pi t}{2} \right) \, dt.$$

Let $$ \begin{align} u&=\frac{\pi t}{2} \\ \Rightarrow du&=\frac{\pi}{2}\, dt \\ \Rightarrow \frac{2}{\pi}\,du &= dt. \end{align} $$

When $t=1$, $$u=\frac{\pi \cdot 1}{2}= \frac{\pi}{2}.$$

When $t=0$,

$$u=\frac{\pi \cdot 0}{2}=0.$$

Hence the substitution,

$$ \begin{align} \int_0^1\cos\left( \frac{\pi t}{2} \right)\, dt &= \frac{2}{\pi}\int_0^{\frac{\pi}{2}}\cos(u)\, du \\ &=\frac{2}{\pi}\bigg[ -\sin(u) \bigg]_0^{\frac{\pi}{2}} \\ &= \frac{2}{\pi}\bigg[ -\sin\left( \frac{\pi}{2} \right)-\left( -\sin(0) \right) \bigg] \\ &=-\frac{2}{\pi}. \end{align} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.