How do you integrate #int x^3 t an x dx # using integration by parts?

Question

How do you integrate #int x^3  t an   x    dx  # using integration by parts?

Charles Anctil · Accepted Answer

#intx^3tan(x)dx = ln|cos(x)|(x^3 - 3x^2 + 6x - 6 + c)#

Say #dv = tan(x)# so #v = ln|cos(x)|# And #u = x^3# so #du = 3x^2#

#intx^3tan(x)dx = x^3*ln|cos(x)| - 3inttan(x)*x^2dx#

For the latter integral, repeat

#dv = tan(x)# so #v = ln|cos(x)|#, #u = x^2# so #du = 2x#

And once more to finish it

#dv = tan(x)# so #v = ln|cos(x)|#, #u = x# so #du = 1#

#intx^3tan(x)dx = x^3ln|cos(x)| - 3x^2ln|cos(x)| + 6(xln|cos(x)| - ln|cos(x)| + c)#

#intx^3tan(x)dx = x^3ln|cos(x)| - 3x^2ln|cos(x)| + 6xln|cos(x)| - 6ln|cos(x)| + c#

Or

#intx^3tan(x)dx = ln|cos(x)|(x^3 - 3x^2 + 6x - 6 + c)#

Samantha Adkison · Answer

undefined To integrate ∫x^3 dx, we can use the integration by parts method. Integration by parts formula is given by: ∫u dv = uv - ∫v du. We choose u and dv such that it simplifies the integral or makes it easier to integrate.

For ∫x^3 dx, we can choose u = x^3 and dv = dx. Then, we find du and v. Taking the derivative of u with respect to x gives du = 3x^2 dx, and integrating dv with respect to x gives v = x.

Now, we substitute u, dv, du, and v into the integration by parts formula:
∫x^3 dx = x^3 * x - ∫(3x^2 * x) dx
= x^4 - ∫3x^3 dx

Now, we have another integral to solve. We can use the integration by parts method again. This time, let's choose u = x^2 and dv = 3x dx. Then, we find du and v. Taking the derivative of u with respect to x gives du = 2x dx, and integrating dv with respect to x gives v = (3/2)x^2.

Now, we substitute u, dv, du, and v into the integration by parts formula:
∫3x^3 dx = x^2 * (3/2)x^2 - ∫(2x * (3/2)x^2) dx
= (3/2)x^4 - ∫3x^3 dx

Now, we have the same integral on both sides of the equation. To solve for the integral, we move it to one side:
∫3x^3 dx = (3/2)x^4 - (3/2)∫3x^3 dx

Now, we can solve for the integral:
∫3x^3 dx = (3/2)x^4 - (3/2) * (3/4)x^4
= (3/2)x^4 - (9/8)x^4
= (3/2 - 9/8)x^4
= (12/8 - 9/8)x^4
= (3/8)x^4

So, the integral of x^3 dx is (3/8)x^4 + C, where C is the constant of integration.