1. ʃ sinx*ln(cosx) dx

First notice the composition ln(cosx). To make the expression easier to integrate we try substitution (substitution is often useful when trying to integrate expressions that are or include compositions of functions).

2. So we let t=cosx.

Now we must find out what dx is in terms of dt. This is quite simple by just differentiating the substitution.

3. dt/dx=-sinx <=> dx=(-1/sinx)dt

Now we substitute everything into our integral from 1.

4. ʃ sinxln(t)(-1/sinx) dt

The sinx from the top and bottom cancel.

5. ʃ -1*ln(t) dt

I have written a -1 since we will have to use integration by parts to integrate ln(t) with respect to t. (Reminder: ʃ u(dv/dx) dx = uv - ʃ v(du/dx) dx)

6. Let u=ln(t) and dv/dt=-1

Now we need to differentiate u to find du/dt and integrate dv/dt to find t.

7. du/dt=1/t and v=-t

Use the intergation by parts formula and that uv=-tln(t) and v(du/dt)=-t(1/t)=-1.

8. ʃ -1*ln(t) dt = -tln(t) - ʃ -1 dt = -tln(t)+t+C

! Remember the constant of integration and that because we used a substitution we must give our answer back in terms of x.

9. So using 2. again we get cosx(1-ln(cosx))+C as the answer.

whats the integral of x.e^x wrt x

Related Maths A Level answers

The quadratic equation 2x^2+ 6x+7 has roots a and b. Write down the value of a+b and the value of ab.

Binomially expand the equation (2+kx)^-3

Integrate sinx*ln(cosx) with respect to x.

Differentiate y = 2x^3 + 6x^2 + 4x + 3 with respect to x.