Question

Solve the following question step by step in detail.
∫²₀ xcosx dx

Answer 1

$\large\int^2_0 xcosx\,dx =$

Using Integration by parts formula,

$\large\int Udv = UV - \int Vdu$

Comparing $\large\int^2_0 xcosx\,dx$ and $\large\int Udv$

$U = x$ and $dv = cosx \,dx$

Since,

$U = x, \quad \dfrac{du}{dx} = 1$

$\therefore du = dx$

$[U = x\,\,, du =dx]$

Since,

$dv = cosx\,dx, \quad \dfrac{dv}{dx}= cosx$

Integrating,

$\therefore V = sinx$

$[V =sinx\,\,, dv = cosx\,dx]$

$\begin{aligned} \end{aligned}$ $\begin{aligned} from, \int Udv &= UV - \int Vdu\\\\ \int^a_b Udv &= {\large[UV]^a_b} - \int^a_b Vdu\\\\ \int^2_0xcosx\,dx &= [xsinx]^a_b- \int^2_0sinx\,dx\\\\ \int^2_0xcosx\,dx &= {\large[xsinx]^2_0}- {\large[-cosx]^2_0}\\\\ \int^2_0xcosx\,dx &= [(2sin(2))-(2sin(0))]- [(-cos(2))-(-cos(0))]\\\\ \int^2_0xcosx\,dx &= [2sin2-(0)]- [-cos2-(-1)]\\\\ \int^2_0xcosx\,dx &= 2sin2- [-cos2+1]\\\\ \int^2_0xcosx\,dx &= 2sin2 +cos2 -1 \end{aligned}$

$\Large\therefore \int^2_0xcosx\,dx\space=\space 2sin2 +cos2 -1 \space \approx \space0.4024$