Subtask 1

Take the integral: ´int sqrt(4-x^2) dx´ For the integrand ´sqrt(4-x^2)´, substitute ´x = 2 sin(u)´ and ´dx = 2 cos(u) du´. Then ´sqrt(4-x^2) = sqrt(4-4 sin^2(u)) = 2 cos(u)´ and ´u = sin^(-1)(x/2)´: ´= 4 int cos^2(u) du´

Write ´cos^2(u)´ as ´1/2 cos(2 u)+1/2´: ´= 4 int (1/2 cos(2 u)+1/2) du´

Integrate the sum term by term and factor out constants: ´= 2 int 1 du+2 int cos(2 u) du´

For the integrand ´cos(2 u)´, substitute ´s = 2 u´ and ´ds = 2 du´: ´= int cos(s) ds+2 int 1 du´

The integral of ´cos(s)´ is ´sin(s)´: ´= sin(s)+2 int 1 du´

The integral of ´1 is u´: ´= sin(s) + 2 u + "constant"´

Substitute back for ´s = 2 u´: ´= 2 u + sin(2 u) + "constant"´

Apply the double angle formula ´sin(2 u) = 2 cos(u) sin(u)´: ´= 2 u + 2 sin(u) cos(u) + "constant"´

Express ´cos(u)´ in terms of ´sin(u)´ using ´cos^2(u) = 1-sin^2(u)´: ´= 2 u + 2 sin(u) sqrt(1-sin^2(u)) + "constant"´

Substitute back for ´u = sin^(-1)(x/2)´: ´= 1/2 sqrt(4-x^2) x + 2 sin^(-1)(x/2) + "constant"´

´int_(-2)^(+2) sqrt(4 − x^2) dx = (1/2 * 0 + 2 sin^(-1)(1) + "constant") - (1/2 * 0 + 2 sin^(-1)(-1) + "constant")´ ´= pi + pi = 2pi´

Subtask 2

´int sin(x) dx = -cos(x) + "constant"´

´int_0^pi sin(x) dx = (-cos(pi) + "constant") - (-cos(0) + "constant") = 1 - (-1) = 2´

Subtask 3

´int e^x x dx´

For the integrand ´e^x x´, integrate by parts, ´int f dg = f g - int g d f´, where ´f = x´, ´dg = e^x dx´, ´df = dx´, ´g = e^x´: ´= e^x x - int e^x dx´

The integral of ´e^x´ is ´e^x´: ´= e^x x - e^x + "constant"´

Which is equal to: ´= e^x (x-1) + "constant"´

´int_0^1 xe^x dx = e^1(1 - 1) - e^0(0 - 1) = 1´