+267 \(\int (2x*sin^{3}x )\, dx\)
I have tried solving it myself by using Integration by parts and I got
\(\frac{6x*cos^{3}(x)-18x*cos(x) -2sin^{3}(x)-12sin(x)}{9}\)
But when I looked the answer up on http://www.integral-calculator.com/ it ended up giving a different answer. I used its "compare an answer" function to see if the two were equivalent but they weren't.
I started by moving the constant 2 outside the integral and tried integrating with u=x which I'd derivate later so it'd become 1.
I ended up having to integrate sin^3x and cos^3x later on which I had to look up online for help with. I tried redoing it to see if I made any small errors in the working out but I got the same answer.
Thanks in advance!
Take the integral:
integral2 x sin^3(x) dx
Factor out constants:
= 2 integral x sin^3(x) dx
For the integrand x sin^3(x), use the trigonometric identity sin^2(x) = 1/2 (1-cos(2 x)):
= integral x sin(x) (1-cos(2 x)) dx
Expanding the integrand x sin(x) (1-cos(2 x)) gives x sin(x)-x sin(x) cos(2 x):
= integral(x sin(x)-x sin(x) cos(2 x)) dx
Integrate the sum term by term and factor out constants:
= - integral x sin(x) cos(2 x) dx+ integral x sin(x) dx
Use the trigonometric identity sin(α) cos(β) = 1/2 (sin(α-β)+sin(α+β)), where α = x and β = 2 x:
= -1/2 integral x (sin(3 x)-sin(x)) dx+ integral x sin(x) dx
Expanding the integrand x (sin(3 x)-sin(x)) gives x sin(3 x)-x sin(x):
= -1/2 integral(x sin(3 x)-x sin(x)) dx+ integral x sin(x) dx
Integrate the sum term by term and factor out constants:
= -1/2 integral x sin(3 x) dx+3/2 integral x sin(x) dx
For the integrand x sin(3 x), integrate by parts, integral f dg = f g- integral g df, where
f = x, dg = sin(3 x) dx, df = dx, g = -1/3 cos(3 x):
= 1/6 x cos(3 x)-1/6 integral cos(3 x) dx+3/2 integral x sin(x) dx
For the integrand cos(3 x), substitute u = 3 x and du = 3 dx:
= 1/6 x cos(3 x)-1/18 integral cos(u) du+3/2 integral x sin(x) dx
The integral of cos(u) is sin(u):
= 1/6 x cos(3 x)-(sin(u))/18+3/2 integral x sin(x) dx
For the integrand x sin(x), integrate by parts, integral f dg = f g- integral g df, where
f = x, dg = sin(x) dx, df = dx, g = -cos(x):
= -3/2 x cos(x)+1/6 x cos(3 x)-(sin(u))/18+3/2 integral cos(x) dx
The integral of cos(x) is sin(x):
= -(sin(u))/18+(3 sin(x))/2-3/2 x cos(x)+1/6 x cos(3 x)+constant
Substitute back for u = 3 x:
= (3 sin(x))/2-1/18 sin(3 x)-3/2 x cos(x)+1/6 x cos(3 x)+constant
Which is equal to:
Answer: |= 1/18 (27 sin(x)-sin(3 x)-27 x cos(x)+3 x cos(3 x))+constant
