Question

$\int _0^{\pi }\sin \left(x\right)dx$Value

Answer 1

$\mathrm{Use\:the\:common\:integral}:\quad \int \sin \left(x\right)dx=-\cos \left(x\right)$

$=\left[-\cos \left(x\right)\right]^{\pi }_0$

$\mathrm{Compute\:the\:boundaries}:\quad \left[-\cos \left(x\right)\right]^{\pi }_0=2$$\int _a^bf\left(x\right)dx=F\left(b\right)-F\left(a\right)=\lim _{x\to \:b-}\left(F\left(x\right)\right)-\lim _{x\to \:a+}\left(F\left(x\right)\right)$

$\lim _{x\to \:0+}\left(-\cos \left(x\right)\right)=-1$

$\lim _{x\to \:\pi -}\left(-\cos \left(x\right)\right)=1$

$=1-\left(-1\right)$

$=2$

Answer 2

Recall that if you have an integral of the form

∫ u(dv/dx) dx

it can be written as

uv – ∫ v(du/dx) dx.

We need to decide which part we will differentiate (as in, which part is u), and which part we will integrate (as in, which part is dv/dx).

We can note that continuously differentiating sin(x) results in a loop of cos(x), –sin(x), –cos(x), sin(x)..., whereas differentiating x once gives 1.

From this, it seems to make sense that we would want to differentiate the x part (so u is x) and therefore integrate the sin(x) part (so dv/dx is sin(x) ). So, let

u = x, which implies du/dx = 1

and let

dv/dx = sin(x). Integrating this to get v gives v = –cos(x).

So our integral is now of the form required for integration by parts.

∫ x sin(x) dx

=  ∫ u(dv/dx) dx

= uv –  ∫ v(du/dx) dx

= –x cos(x) –  ∫ –cos(x)*1 dx

= –x cos(x) –  ∫ –cos(x) dx

= –x cos(x) +  ∫ cos(x) dx