# How do you integrate int x+cosx from [pi/3, pi/2]?

May 4, 2018

The answer ${\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} x + \cos x \cdot \mathrm{dx} = 0.8193637907356557$

#### Explanation:

show below

${\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} x + \cos x \cdot \mathrm{dx} = {\left[\frac{1}{2} {x}^{2} + \sin x\right]}_{\frac{\pi}{3}}^{\frac{\pi}{2}}$

$\left[{\pi}^{2} / 8 + \sin \left(\frac{\pi}{2}\right)\right] - \left[{\pi}^{2} / 18 + \sin \left(\frac{\pi}{3}\right)\right] = \frac{5 \cdot {\pi}^{2} - 4 \cdot {3}^{\frac{5}{2}} + 72}{72} = 0.8193637907356557$

May 4, 2018

${\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} \left(x + \cos x\right) \mathrm{dx} = 1 + \frac{5 {\pi}^{2} - 36 \sqrt{3}}{72}$

#### Explanation:

Using the linearity of the integral:

${\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} \left(x + \cos x\right) \mathrm{dx} = {\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} x \mathrm{dx} + {\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} \cos x \mathrm{dx}$

Now:

${\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} x \mathrm{dx} = {\left[{x}^{2} / 2\right]}_{\frac{\pi}{3}}^{\frac{\pi}{2}} = {\pi}^{2} / 8 - {\pi}^{2} / 18 = \frac{5 {\pi}^{2}}{72}$

${\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} \cos x \mathrm{dx} = {\left[\sin x\right]}_{\frac{\pi}{3}}^{\frac{\pi}{2}} = \sin \left(\frac{\pi}{2}\right) - \sin \left(\frac{\pi}{3}\right) = 1 - \frac{\sqrt{3}}{2}$

Then:

${\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} \left(x + \cos x\right) \mathrm{dx} = 1 + \frac{5 {\pi}^{2} - 36 \sqrt{3}}{72}$

May 4, 2018

(5π^2)/72+1-sqrt3/2

#### Explanation:

int_(π/3)^(π/2)(x+cosx)dx $=$

int_(π/3)^(π/2)xdx+int_(π/3)^(π/2)cosxdx $=$

[x^2/2]_(π/3)^(π/2) $+$ [sinx]_(pi/3)^(π/2) $=$

(π^2/4)/2-(π^2/9)/2+sin(π/2)-sin(π/3) $=$

π^2/8-π^2/18+1-sqrt3/2 $=$

(5π^2)/72+1-sqrt3/2