Some more integrals please help

2) Consider the following integrals

a) Find ʃ x^2 lnx dx

b) ʃ e^x sin x dx

Please show me the working out so i know what has been done
Thank you so much :-)

Integration by parts is much easier to do in terms of functions and derivatives rather than differentials, yet the below answerer has chosen the more complicated way.

a.
Integrate the original integrand by parts:
∫ x²lnx dx
Let f'(x) = x²
f(x) = x³ / 3
Let g(x) = lnx
g'(x) = 1 / x
∫ f'(x)g(x) dx = f(x)g(x) - ∫ f(x)g'(x) dx
∫ x²lnx dx = x³lnx / 3 - ∫ x² / 3 dx
∫ x²lnx dx = x³lnx / 3 - x³ / 9 + C
∫ x²lnx dx = x³(3lnx - 1) / 9 + C

b.
Integrate the original integrand by parts:
∫ ℮˟sinx dx
Let f'(x) = ℮˟
f(x) = ℮˟
Let g(x) = sinx
g'(x) = cosx
∫ f'(x)g(x) dx = f(x)g(x) - ∫ f(x)g'(x) dx
∫ ℮˟sinx dx = ℮˟sinx - ∫ ℮˟cosx dx

Integrate the new integrand by parts:
∫ ℮˟cosx dx
Let f'(x) = ℮˟
f(x) = ℮˟
Let g(x) = cosx
g'(x) = -sinx
∫ f'(x)g(x) dx = f(x)g(x) - ∫ f(x)g'(x) dx
∫ ℮˟cosx dx = ℮˟cosx + ∫ ℮˟sinx dx

Put it all together to solve for the original integral:
∫ ℮˟sinx dx = ℮˟sinx - ∫ ℮˟cosx dx
∫ ℮˟sinx dx = ℮˟sinx - (℮˟cosx + ∫ ℮˟sinx dx)
∫ ℮˟sinx dx = ℮˟sinx - ℮˟cosx - ∫ ℮˟sinx dx
2 ∫ ℮˟sinx dx = ℮˟sinx - ℮˟cosx
2 ∫ ℮˟sinx dx = ℮˟(sinx - cosx)
∫ ℮˟sinx dx = ℮˟(sinx - cosx) / 2 + C

2.) ʃ e^x sin x dx


Use Integration by parts

Let u = e^x and dv = sin x dx


Integral [ u dv] = uv - vdu -> This is integration by parts

ʃ e^x sin x dx = - e^x * cos(x) + ʃ e^x * cos(x) dx


Now for the integral ʃ e^x * cos(x) dx

Let s = e^x and dt = cos(x) dx

Using Integrating by parts we get

ʃ e^x * cos(x) dx = e^x * sin(x) - ʃ e^x * sin(x) dx

Now substitute this expression into the following equation

ʃ e^x sin x dx = -e^x * cos(x) + ʃ e^x * cos(x) dx

and we get

ʃ e^x sin x dx = -e^x * cos(x) + [e^x * sin(x) - ʃ e^x * sin(x) dx]


Now add ʃ e^x * sin(x) dx to the left side of the equation to get

2 * ʃ e^x sin x dx = e^x * sin(x) - e^x * cos(x)


Now simplify to get

ʃ e^x sin x dx = [e^x * sin(x) - e^x * cos(x)] / 2 + C