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

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

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

$$= \frac 2 \pi$$

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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?

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HINT:

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

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

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

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