Problem with integraton by parts?

in the first cases you cited, you can look at the x^3 and x^2 term and recognize that if you set u=x^3, the first integration by parts will give you a term du (3x^2) e^x; the second int by parts will give 6x e^x, and the third will have 6e^x.. so in these cases, the power of x is thenumber of times you have to integrate by parts

in the case of e^x sinx, you set I= Integral (e^x sin x dx) and integrate by parts

first set u=e^x so du = e^x dx; dv=sin x so v=-cos x

then you get

I = -cos x e^x -Integral[-cos x e^x dx]

integrate by parts again, getting:

I=-cos x e^x+[sinx e^x-Integral[e^x sinx dx]

and it may look like you are in an infinite loop, but notice that the final integral on the right is just the initial integral, I, so treating the integral as an algebraic expression, you get:

2I = -cos x e^x+sin x e^x or I = (1/2) e^x(sinx -cosx)

For bothese problems, I suggest to you simpler methods.

For a problem of the type, you can straightaway write the answer as under.

∫e^x * x^3 dx = (x^3 - 3x^2 + 6x - 6) e^x + c.

What I have done is to write in the bracket the given polynomial, here x^3, then change the sign and write its differentiation, change the sign and write its second differentiation and so on till the last term is a constant. Complete the bracket and multiply by e^x gives the answer.

Thus, ∫f (x) * e^x dx, where f(x) is a polynomial in x of any degree

= [f (x) - f '(x) + f "(x) - f"'(x) + .... till a constant term] e^x + c.

Take one more example, ∫(x^3 + 2x^2 + 3x + 4) e^x dx

= [(x^3+2x^2+3x+4) - (3x^2+4x+3) + (6x+4) - 6] e^x + c

= (x^3 - x^2 + 5x - 5) e^x + c.

[The nuber of repeated integration needed = degree of the polynomial.]

For the second type of integration, ∫ e^2x * sin (x) dx, use the formula

∫e^(ax) * sin(bx + k) dx = [e^ax/(a^2+b^2)] * (asin(bx+k) - bcos(bx+k)] + c.

[Note: This integration has to be done twice by parts as sin changes to cos first time and back to sin second time.]

[For derivation of this formula, you may refer to my free educational website of math and physics which I have created as a part of my hobby. The first link takes you to the derivation of the formula and the second link is of the homepage of the website.]

1) For e^x * x^3, three IBPs are not necessary - proof:

let u = e^x

du = e^x dx

Let dv = x^3 dx

v = x^4 / 4 dx

Then ∫e^x * x^3 dx = e^x * (x^4 / 4) - ∫ e^x * (x^4 / 4) dx

Let u = (x^4 / 4)

Then du = (3x^3 / 4) dx

Let dv = e^x dx

Then v = e^x

So: Then ∫e^x * x^3 dx = e^x * (x^4 / 4) - (((3x^4 * e^x)) / 4) + ∫(3x^3*e^x) dx / 4

Subtract the right most expression from both sides of the equation, getting:

(1/4) * ∫e^x * x^3 dx = e^x * (x^4 / 4) - (3x^3 * e^x / 4).............and, multiplying both sides by 4:

∫e^x * x^3 dx = (4 * [e^x * (x^4) - (3x^3 * e^x] )

Three IBPs are clearly not necessary. This is true for e^x * x^n for any positive integer n and I suspect every real positive number n.

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This same technique works for your problem and for:

∫ e^(7x) * sin(x) dx = (7 * e^(7x)) * [(x/7) - (1/49)]

Two steps almost always. I've never seen an exception to the rule.

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