Question

limit 0 to Pi /2 integration xsinx

Answer 1

Answer:

$\bf\int\limits^\frac{\pi}{2}_0{x\;sinx}\;dx=1$

Step-by-step explanation:

I have applied bernoulli's formula to evaluate this integral

$\textbf{Bernoulli's formula:}$

$\boxed{\bf\int{u}\;dv=u\;v-u'\;v_1}$

$Let\;\;I=\int\limits^\frac{\pi}{2}_0{x\;sinx}\;dx$

Take

$u=x$

$u'=1$

and

$dv=sinx\;dx$

$\int{dv}=\int{sinx}\;dx$

$v=-cosx$

$v_1=-sinx$

Now, Bernoulli's formula

$I=[-x\;cosx+1\,sinx]\limits^\frac{\pi}{2}_0$

$I=[-\frac{\pi}{2}\;cos\frac{\pi}{2}+sin\frac{\pi}{2}]-[-0cos0+sin0]$

$I=-\frac{\pi}{4}(0)+1$

$\implies\bf\,I=1$