Help Solve The Integral

Help, How do I solve this integral?

let integral of e^8x cos5x dx be called A
then by integration by parts
A=e^8xsin5x/5 - 8/5 integral of(e^8x sin5x)dx
again integrating integral of(e^8x sin5x)dx by parts
=e^8xsin5x/5- 8/5 {e^8 (-cos5x/5) + 8/5 integral of e^8x cos5x dx }

A =e^8xsin5x/5 + 8/25 e^8xcos5x - 64/25 A
A+64/25A = e^8xsin5x/5 + 8/25 e^8xcos5x

89/25 A= e^8xsin5x/5 + 8/25 e^8xcos5x
89/5 A=e^8xsin5x + 8/5 e^8xcos5x
A= 5/89 { e^8xsin5x + 8/5 e^8xcos5x }

which was the required answer

hope this helped.....

Solve the integral..?

E) None of these

-3/2 cos^2 (x) + const.

or

3/2 sin^2 (x) + const

You could use u type substitution, let u=sin x, then du = cos x dx

Then the integral becomes integral of 3 u du = 3u^2/2 + const
=3/2 sin^2 x + const.

The other similarly using u= cosx, du = -sinx dx...

Both answers are the same since cos^2= 1 - sin^2, and the extra 3/2 you'll get changing one into the other form will be absorbed by the const of integration. Another way to say it, the derivative of either answer is 3 cos(x) sin(x)

How Do I solve this integral?

I plugged the integral into mathcad and symbolically integrated it to get: -1/(x-2)^2+1/(x-2). This is an integration by parts type of problem. Look up this procedure in you calc book. Evaluate this with x=1 and subtract the quantity with x=0 and you get -1/4.

You could do this numerically by drawing the curve on graph paper (each box = 0.1 or 0.05) and then count the boxes (each has area of 0.1^2 or 0.05^2). This is how old people used to numerically integrate.

Help solving this integral!?

You can differentiate the whole thing with respect to x to get a differential equation, which is solvable. When you differentiate the integral, use Leibniz's differentiation rule since you have to be more careful (you have variable limits in your integral rather than constants). Leibniz's rule is something you probably were never told about, but you are intended to figure at some point that it exists, usually in the course of doing homework for classes (that is how most of us found out about it).

You can solve the differential equation using methods you all ready studied, now how to find the integration constant? Use their hint, the initial condition can be obtained from the integral equation:

y(x) = 49 + ∫2t*sqrt(y(t)) dt from 0 to x

the initial condition happens at x = 0:

y(0) = 49 + ∫2t*sqrt(y(t)) dt from 0 to 0 --> y(0) = 49 since any integral evaluated over the same point is zero (there is zero area under such a curve).

Make sense?

Help solving an integral?

Let I(n) = ∫[0,1] x^n e^x dx. Let u = x^n, dv = e^x dx. Then du = n x^(n - 1) dx and v = e^x. By integration by parts,

I(n) = (x^n)(e^x)|[x = 0 to 1] - ∫[0,1] (e^x)(n x^(n - 1)) dx = e - n I(n - 1).

Reiterating the recursion,

I(n) = e - n[e - (n - 1) I(n - 2)] = e - ne + n(n - 1) I(n - 2)

= e - ne + n(n - 1)[e - (n - 2) I(n - 3)] = e - ne + n(n - 1) e - n(n - 1)(n - 2) I(n - 3)

= e - ne + n(n - 1) e - n(n - 1)(n - 2)[e - (n - 3) I(n - 4)]

= e - ne + n(n - 1) e - n(n - 1)(n - 2) e + n(n - 1)(n - 2)(n - 3) I(n - 4)
.
.
.
= e - ne + n(n - 1) e - n(n - 1)(n - 2) e + ... + (-1)^k n(n - 1)...(n - k + 2) e + ...
+ (-1)^n n! I(0).

Since I(0) = ∫[0,1] e^x dx = e, the result is

I(n) = e - ne + n(n - 1) e - n(n - 1)(n - 2) e + ... (-1)^k n(n - 1)... (n - k + 1) e + ...
+ (-1)^n n! e,

= e[1 - n + n(n - 1) - n(n - 1)(n - 2) + ... + (-1)^k n(n - 1)...(n - k + 1) + ... + (-1)^n n!]

= e[1 - 1! (n 1) + 2! (n 2) - 3! (n 3) + ... + (-1)^k k! (n k) + ... + (-1)^n n!],

where the (n k) = n!/k!(n - k)! are the binomial coefficients.

How do i solve integrals?

For this one, you can multiply the x in to the parenthesizes to get x + x^4

You then have:
∫ (x + x^4) dx

From that, because there is just addition between them (no multiplication), you can find the integral in two parts:
∫ (x) dx + ∫ (x^4) dx

For this, just think of the integral as the anti-derivative, the derivative of what would give x? well, using the power rule (the derivative of x^n = (n)(x^(n-1))) you might notice that you could raise the power of the x to 2, then if you were to find the derivative of x^2, you would get 2x, but we want 1x, so if we had (2/2)x, it would be 1x, so the derivative of (1/2)(x^2) would be just x, thus the integral of x (opposite of derivative) would be (1/2)(x^2):
(1/2)(x^2) + ∫ (x^4) dx

You can then do the same thing for the next one...the derivative of what would give x^4, well, using the power rule again, if we had x^5, when we subtracted one from the power, we would get x^4, but then multiplied it by 5, and have 5x^4, but if it was (1/5)(x^5), when we found the derivative, it would be (5/5)(x^4), or x^4, so the integral would be (1/5)(x^5)
(1/2)(x^2) + (1/5)(x^5)

That would be the answer...or you could write it so it is all one fraction, instead of the sum of two.
((x^2) / 2) + ((x^5) / 5)
((5x^2) / 10) + ((2x^5) / 10)
(5x^2 + 2x^5) / 10

Help solving integral...?

5 - x^2 = u
-2x * dx = du

x^5 * (5 - x^2)^(1/2) * dx =>
x^2 * x^2 * (5 - x^2)^(1/2) * x * dx =>
(-1/2) * x^2 * x^2 * (5 - x^2)^(1/2) * (-2 * x * dx) =>
(-1/2) * (5 - u) * (5 - u) * u^(1/2) * du =>
(-1/2) * (25 - 10u + u^2) * u^(1/2) * du =>
(-1/2) * (25 * u^(1/2) - 10 * u^(3/2) + u^(5/2)) * du

Now it's just a matter of using the power rule to integrate:

(-1/2) * (25 * (2/3) * u^(3/2) - 10 * (2/5) * u^(5/2) + (2/7) * u^(7/2)) + C
(-1/2) * 2 * u^(3/2) * ((25/3) - (10/5) * u + (1/7) * u^2) + C
-1 * u^(3/2) * ((25/3) - 2u + (1/7) * u^2) + C
-u^(3/2) * (1/3) * (1/7) * (25 * 7 - 2 * 3 * 7 * u + 3 * u^2) + C
-u^(3/2) * (1/21) * (175 - 42u + 3u^2) + C
(-1/21) * (3u^2 - 42u + 175) * u^(3/2) + C
(-1/21) * (3 * (5 - x^2)^2 - 42 * (5 - x^2) + 175) * (5 - x^2)^(3/2) + C
(-1/21) * (3 * (25 - 10x^2 + x^4) - 210 + 42x^2 + 175) * (5 - x^2)^(3/2) + C
(-1/21) * (75 - 30x^2 + 3x^4 - 210 + 42x^2 + 175) * (5 - x^2)^(3/2) + C
(-1/21) * (3x^4 + 12x^2 + 40) * (5 - x^2)^(3/2) + C

That's about as nice as it's going to get.