Find the integral of x^3 + [(x^4)(tanx)] from -pi/4 to pi/4
In exams, watch out for these freebies, definite integrals only!
If you can show that the function to be integrated is odd (i.e. f(x)=-f(x), and the limits of integration is symmetric around 0, e.g. -%pi/4 to %pi/4, etc., the answer is zero!
Try that with known odd functions, ex. y=x, y=x^3, y=sin(x), y=tan(x), etc.
When it is a product of an even function (x^2) and an odd function (sin(x)), the result is still odd: example:
(x^2)tan(x) is odd, so
∫x^2 tan(x)dx =0 if integrated from -π/4 to π/4.
But DO SHOW that the function is odd and the limits of integration are symmetric around zero before making the conclusion.
f(x)=-f(-x) ∀x => odd function
To find the integral of the given function, we can use integration by parts. The formula for integration by parts is:
∫(u * v) dx = u * ∫v dx - ∫(u' * ∫v dx) dx
Let's break down the given function and assign u and v:
u = x^3 ---> u' = 3x^2
v = ∫[(x^4)(tanx)] dx
Now, let's compute v using another integration technique, which is integration by substitution.
Let z = x^4
dz = 4x^3 dx
dx = dz / (4x^3)
Substituting this into the integral of v, we have:
v = ∫(z * tanx) (dz / (4x^3))
= ∫(z * tanx) dz / (4x^3)
= ∫(z * tanx) / (4x^3) dz
To integrate ∫(z * tanx) / (4x^3) dz, we can use another integration technique called integration by parts. Assigning u and v:
u = tanx ----> u' = sec^2x
v = z / (4x^3) ---> v' = 1 / (4x^3)
Using the integration by parts formula:
∫(u * v') dx = u * ∫v' dx - ∫(u' * ∫v' dx) dx
Let's compute the integration:
∫(u * v') dz = u * ∫v' dz - ∫(u' * ∫v' dz) dz
∫(tanx * (z / (4x^3))) dz = tanx * ∫(1 / (4x^3)) dz - ∫((sec^2x) * (∫(1 / (4x^3)) dz)) dz
Simplifying further:
∫(tanx * (z / (4x^3))) dz = (1 / (4x^3)) * tanx * z - ∫((sec^2x) * (1 / (4x^3))) dz
Now we have the integral of v. Let's substitute this back into the original formula for integration by parts:
∫(u * v) dx = u * ∫v dx - ∫(u' * ∫v dx) dx
∫((x^3) + [(x^4)(tanx)]) dx = (x^3) * ∫(1 / (4x^3)) * tanx * z dz - ∫(3x^2) * (1 / (4x^3)) * tanx * z dz dx
We can now simplify and evaluate this definite integral from -π/4 to π/4.