How do you integrate #int x^3 cos^2x dx # using integration by parts?

Question

How do you integrate #int x^3  cos^2x    dx  # using integration by parts?

Charles Anctil · Accepted Answer

# = x^4/8 + x^3 /4 sin2x + (3)/8 x^2 cos 2x - 3/8 x sin 2x - 3/16 cos2x + C#

that's very ugly and would need a load of IBP so i would start by making it even bigger!! using #cos 2A = 2 cos^2 A - 1#

#I = int x^3 cos^2x \ dx#

#= int x^3 (cos 2x +1 )/2 \ dx#

#implies 2I = int x^3 cos 2x +x^3 \ dx#

#implies 2I - x^4/4 = int x^3 cos 2x \ dx qquad triangle# ......taking the easy bit to the LHS

so now we IBP the remaining bit over and over, reducing the #x^3# term each time

# int \ x^3 cos 2x \ dx = int \ x^3 d/dx (1/2 sin 2x) \ dx#

which by IBP.... #= x^3 /2 sin2x - int \ 3x^2 1/2 sin 2x \ dx#

#= x^3 /2 sin2x - (3)/2 int \ x^2 sin 2x \ dx#

#= x^3 /2 sin2x - (3)/2 int \ x^2 d/dx( - 1/2 cos 2x) \ dx#

#= x^3 /2 sin2x + (3)/4 int \ x^2 d/dx( cos 2x) \ dx#

#= x^3 /2 sin2x + (3)/4 ( x^2 cos 2x - int \ 2x cos 2x \ dx) #

and on we go!!

#= x^3 /2 sin2x + (3)/4 ( x^2 cos 2x - int \ 2x d/dx (1/2 sin 2x ) \ dx) #

#= x^3 /2 sin2x + (3)/4 ( x^2 cos 2x - int \ x d/dx ( sin 2x ) \ dx) #

#= x^3 /2 sin2x + (3)/4 ( x^2 cos 2x - (x sin 2x - int \ sin 2x \ dx)) #

#= x^3 /2 sin2x + (3)/4 ( x^2 cos 2x - (x sin 2x + 1/2 cos2x )) + C#

#= x^3 /2 sin2x + (3)/4 x^2 cos 2x - 3/4 x sin 2x - 3/8 cos2x + C#

so back to #triangle#

#implies 2I - x^4/4 = x^3 /2 sin2x + (3)/4 x^2 cos 2x - 3/4 x sin 2x - 3/8 cos2x + C#

#implies I = x^4/8 + x^3 /4 sin2x + (3)/8 x^2 cos 2x - 3/8 x sin 2x - 3/16 cos2x + C#

once again , the key is IBP:

#int \ u(x) * d/dx(v(x)) \ dx = u(x) * v(x) - int \ d/dx(u(x)) * v(x) \ dx#

Paisley Johnson · Answer

undefined To integrate $ \int x^3 \cos^2(x) \, dx $ using integration by parts:

Let $ u = x^3 $ and $ dv = \cos^2(x) \, dx $.
Then, $ du = 3x^2 \, dx $ and $ v = \frac{x}{2} + \frac{\sin(2x)}{4} $.

Using the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

We get:

$$ \int x^3 \cos^2(x) \, dx = \frac{x^3}{2} + \frac{x\sin(2x)}{4} - \int \left(\frac{x}{2} + \frac{\sin(2x)}{4}\right)3x^2 \, dx $$

$$ = \frac{x^3}{2} + \frac{x\sin(2x)}{4} - \frac{3}{2}\int x^3 \, dx - \frac{3}{4}\int \sin(2x) \, dx $$

$$ = \frac{x^3}{2} + \frac{x\sin(2x)}{4} - \frac{3}{8}x^4 - \frac{3}{8} \cos(2x) + C $$

Where $ C $ is the constant of integration.