Please can anyone help with the following problems - thanks.
1) Integrate X^4 e^x dx
2) Integrate Cos^5(x) dx
3) Integrate Cos^n(x) dx
4) Integrate e^(ax)Sinbx dx
5) Integrate 5xCos3x dx
The standard way to solve most of these integrals is using partial integration. So, look up partial integration in your book or on the web.
Perhaps my time here is better spend by demonstrating how you can solve all these integrals using an alternative method that is often much faster.
1) Integrate x^4 e^x dx
I would do this as follows. Instead of x^4 e^x integrate e^(a x):
Integral Exp(a x) dx = Exp[a x]/a
And now we differentiate both sides four times w.r.t. a. You can interchange integration over x and differentiation w.r.t. a to obtain:
Integral x^4 Exp(a x) dx =
[24/a^5 - 24x/a^4 + 12 x^2/a^3
-4 x^3/a^2 + x^4/a] Exp(a x)
All you have to do now is put a = 1. Note that we don't bother about the integration constant here, you can add that to the result you obtain the end.
The integration constant you could have added at the beginning doesn't drop out when you differentiate w.r.t. a, because the most general integration constant is an arbitrary function of a.
2) Integrate Cos^5(x) dx
We write:
Cos^5(x) = cos(x) (1-sin^2(x))^2=
Substitute sin(x) = y, then
cos(x)dx = dy
The integral becomes:
∫ cos^5(x) dx = ∫ (1-y^2)^2 dy
Using the power rule for integration, we can expand the expression:
= ∫ (1-2y^2 + y^4) dy
Integrating term by term, we get:
= y - (2/3)y^3 + (1/5)y^5 + C
Now, substitute y back as sin(x):
= sin(x) - (2/3)sin^3(x) + (1/5)sin^5(x) + C
3) Integrate Cos^n(x) dx
Similar to the previous problem, we can expand the expression using the binomial theorem:
Cos^n(x) = (cos(x))^n = (1-sin^2(x))^n
Using the binomial theorem, we can expand this as:
Cos^n(x) = 1 - n sin^2(x) + (n(n-1)/2)sin^4(x) - (n(n-1)(n-2)/6)sin^6(x) + ...
Integrating term by term, we get:
= x - (n/3)x^3 + (n(n-1)/10)x^5 - (n(n-1)(n-2)/42)x^7 + ... + C
4) Integrate e^(ax)Sin(bx) dx
For this integral, we can use integration by parts. The formula for integration by parts is:
∫ u dv = uv - ∫ v du
Choosing u = e^(ax) and dv = sin(bx) dx, we get:
du = a e^(ax) dx and v = - (1/b) cos(bx)
Applying the formula, we have:
∫ e^(ax)sin(bx) dx = - (1/b) e^(ax) cos(bx) + (a/b) ∫ e^(ax) cos(bx) dx
Now, we have another integral of the form e^(ax) cos(bx). To solve it, we can use integration by parts again.
Choosing u = e^(ax) and dv = cos(bx) dx, we get:
du = a e^(ax) dx and v = (1/b) sin(bx)
Applying the formula, we have:
∫ e^(ax) cos(bx) dx = (1/b) e^(ax) sin(bx) - (a/b) ∫ e^(ax) sin(bx) dx
Substituting this back into the original integral, we get:
∫ e^(ax)sin(bx) dx = - (1/b) e^(ax) cos(bx) + (a/b) [(1/b) e^(ax) sin(bx) - (a/b) ∫ e^(ax) sin(bx) dx]
Simplifying, we get:
∫ e^(ax)sin(bx) dx = - (1/b) e^(ax) cos(bx) + (a/b^2) e^(ax) sin(bx) - (a^2/b^2) ∫ e^(ax) sin(bx) dx
Now, we can solve this equation for the integral:
[(a^2/b^2) + 1] ∫ e^(ax)sin(bx) dx = - (1/b) e^(ax) cos(bx) + (a/b^2) e^(ax) sin(bx)
Dividing both sides by [(a^2/b^2) + 1], we get:
∫ e^(ax)sin(bx) dx = - (1/b) [(a/b) e^(ax) cos(bx) - e^(ax) sin(bx)] / [(a^2/b^2) + 1]
5) Integrate 5xCos(3x) dx
For this integral, we can use integration by parts again. Choosing u = 5x and dv = cos(3x) dx, we get:
du = 5 dx and v = (1/3) sin(3x)
Applying the formula, we have:
∫ 5xCos(3x) dx = (5/3) x sin(3x) - (5/3) ∫ sin(3x) dx
Using the formula for integrating sin(x), we get:
∫ 5xCos(3x) dx = (5/3) x sin(3x) + (5/9) cos(3x) + C
And that's the solution to the integral.