1. ## Integ.

Hello,
sorry i cant learn latex
0 to (2sgrt3)S(sgrt(16-x^2)-2)dx=?
S=integ.

i want to see solution way then i can solve ...Thanks for all help

2. Originally Posted by bebrave
Hello,
sorry i cant learn latex
0 to (2sgrt3)S(sgrt(16-x^2)-2)dx=?
S=integ.

i want to see solution way then i can solve ...Thanks for all help

$\int_0^{2\sqrt{3}}\left(\sqrt{16-x^2} - 2\right)dx=\int_0^{2\sqrt{3}}\left(\dfrac{16-x^2}{\sqrt{16-x^2}} - 2\right)dx =$ $\int_0^{2\sqrt{3}}\left(\dfrac{16}{\sqrt{16-x^2}}-\dfrac{\frac12 x \cdot 2x}{\sqrt{16-x^2}} - 2\right)dx =$

The first quotient will give an arcsin-function, the second quotient has to be integrated by partial integration and integration by substitution.

3. i think this way,
x=4sint-->dx=4cost.dt
x=0==>sint=0==>t=0
x=2sgrt3==>sint=(sgrt3)/2==>t=x/3
so
0 to pi/3 S(sgrt(16-16sin^2t)-2)4cost.dt
0 to pi/3 S(4cost-2)4cost.dt
0 to pi/3 S(16cos^2t-8cost)dt
then