Integral questions?

Question

Plz write the steps to get to the answer, as well as the method of integration you used. Also if any part needs some thing from a law or rule, the the fundamental theorom of calc, plz mention that u used it and where:

int() = integral ()

1. int(x^3 * e^(-x^2) dx)
2. int(x^3 * (2+x^2)^(5/2) dx
3. int( abs(cos^2(x) * sin(x)) dx) range of integral is 0 to 2 pi and abs is absolute value of
4. int(x*e^(sin(x^2))) dx) range is -1 pi to +1 pi and i know the answer is 0 because the function iside the int is like oscilating...forgot the term. but plz either show how you prove its is even(maybe thats the term) or plz solve the integral.

Tks a bunch. Just write the number of the question by your answer. tks again

Answer 1

1. Use integration by parts: Let u = x^2 and
dv = (e^(-x^2))xdx.
Then du = 2xdx and v = -(e^(-x^2))/2. Let I = the assigned integral. You have
I = -(x^2)(e^(-x^2)) + int(x(e^(-x^2))dx
= " + (-1/2)(e^(-x^2)) + C

2. Again, use integration by parts. Let u = x^2 and dv = x(2 + x^2)^(5/2). Then du =2xdx and v = (1/7)*(2 + x^2)^(7/2). Now
I = (1/7)*(x^2)*(2 + x^2)^(7/2) - (1/7)*int((2x)(2 + x^2)&(7/2)
= " - (1/7)*[(2 + x^2)^(9/2)]*(2/9) + C.

3. First notice this: on [0,pi] abs(((cos x)^2 )*sin x) =
((cos x)^2)*sin x but on [pi,2pi] the abs = -((cos x)^2)*sin x.
Let I[a,b] denote the definite integral on [a,b] . Then
I[0,pi] = -(1/3)*((cos x)^3) = -(1/3)*[(cos pi)^3 - (cos 0)^3]
= -(1/3)[(-1)^3 - 1^3] = 2/3. Also,
I[pi,2pi] = (1/3)*(cos x)^3 = (1/3)[(cos 2pi)^3 - (cos pi)^3]
= (1/3)*[1^3 - (-1)^3] = 2/3.
Therefore, I[0,2pi] = 4/3.

4. I think the word you are looking for is "odd." A function f is odd if and only if f(-x) = -f(x). Let f(x) = x*e^(sin(x^2)) Then
f(-x) = (-x)*e^(sin((-x)^2)) = -f(x). Then I[-pi,0] = -I[0,pi] so as you noticed, I[-pi,pi] = 0.

Answer 2

1)
Let u = -x²
du = -2xdx <=> xdx = -du/2

You can factor x^3 as x² * x (Get 1 x from here)

So, the resulting integral would be

∫-u e^u (-du/2) (negative sign will just cancel, you take 1/2 outside the integral)

(1/2)∫u e^u du (Use integration by parts: ∫u dv = uv - ∫vdu, i just changed the variables since i already used u)

Let v = u dw = e^u
dv = du w = e^u

Applying the formula bove:

(1/2)(u e^u - ∫e^u du)

(1/2)(u e^u - e^u) + C

Substituting back to x:

(1/2)[(-x² e^(-x²)) - e^(-x²)] + C

2)
∫x^3 (2 + x²)^5/2 dx

Let u = 2 + x² <=> x² = u - 2
du = 2xdx <=> xdx = du/2

Again, x^3 = x² * x

So we will end up with...

∫(u-2) u^5/2 du/2 (Again, 1/2 will just go out of the integral, distribute u^5/2, keep in mnind the laws of exponents)

(1/2)∫u^7/2 - 2u^5/2 du (Integral can be distributed over difference)

(1/2)∫u^7/2 du - (1/2)∫2u^5/2 du (Integral of a power
∫x^n = [x^(n+1)]/n+1)

(1/2)(2/9)u^9/2 - (2/7)u^7/2 + C (I already simplified)

Going back to x...

(1/9)(2 - x²)^9/2 - (2/7)(2 + x²)^7/2 + C

I guess you can do the rest, keep in mind
∫|x|dx = ∫xdx = x²/2 + C