Calculus: I Need Help With Some Integration Problems?

Question

I worked out the following problem by first integrating by parts, and then solving the second integral via trig substitution. After combining everything together, I got pi/288 as my answer, but the fnint function (which can calculate the numerical value of a definite integral) on my calculator is giving me a different answer. Please work out this problem step-by-step so I can see exactly how you are supposed to do it:


1. x arcsin(6x) dx from 0 to 1/6 (definite integral)

(which reads as x times the sine inverse of 6x).


I tried doing a U substitution for this problem, but I’m not exactly sure how you are supposed to do it, or if you have to do multiple substitutions before you can integrate. Please work out this problem step-by-step so I can see the correct approach:

2. x^2 (e^x)^(1/4) dx from 0 to 1 (definite integral)

(which reads as x squared times the fourth root of e^x; note that the entire expression of e^x is under the fourth root).

Answer 1

u ... inv.trigo/logarithmic ... polynomial ... trigo/exponential... dv

the above sequence will give you a good guide of how to approach integration by parts... this means if you have a logarithmic and a polynomial... the logarithmic will go to U and the polynomial will go to dV... but that is just a general guide. §

For your first problem:
u = arcsin(6x) ... dv = x dx
du = 6dx/√(1-36x²) ... v = x² / 2

By IBP, ∫ u dv = uv - ∫ v du ...... Thus,
∫ x arcsin(6x) dx = 1/2 x² arcsin(6x) - ∫ 3x²/√(1-36x²) dx

Now, to solve ∫ 3x²/√(1-36x²) dx , we let 6x = sinθ ... 6 dx = cosθ dθ
The question becomes...
∫ 3sin²θ/ {36*√(1-sin²θ)} [1/6 cosθ dθ]
= 1/72 ∫ sin²θ dθ ... by the trig. identity
= 1/72 ∫ [1 - cos2θ]/2 dθ
= 1/144 (θ - 1/2 sin2θ ) ... by double angle formula
= 1/144 (θ - sinθ cosθ)
= 1/144 (arcsin(6x) - (6x)√(1-36x²))

Thus, ∫ x arcsin(6x) dx =
1/2 x² arcsin(6x) - 1/144 arcsin(6x) + 1/24 x√(1-36x²) ... evaluated from 0 to 1/6
[note arcsin 1 =π/2]
= 1/2 * 1/36 * π/2 - π/288 + 1/24*1/6*0
= π/144 - π/288
= π/288


2. This time u = x² ... dv = e^(x/4) dx
... du = 2x dx ... v = 4e^(x/4)

∫ x^2 e^(x/4) dx ...
= 4x²e^(x/4) - 8 ∫ x e^(x/4) dx
(and by another IBP)
= 4x²e^(x/4) - 8( 4xe^(x/4) - 16e^(x/4) )
... this time from 0 to 1 ...
= 4e^.25 - 32e^.25 + 128e^.25 - 128e
= 100 e^.25 - 128

Answer 2

#1: The first step is to integrate by parts, and then integrate the second part using a trig substitution.

[0, 1/6]∫x arcsin (6x) dx

First, integrate by parts: let u = arcsin (6x), du = 6/√(1-(6x)²) dx, dv = x dx, v = x²/2. Then we have:

1/2 x² arcsin (6x) |[0, 1/6] - [0, 1/6]∫3x²/√(1-36x²) dx

Evaluating the first part:

1/2 (1/36) arcsin (1) - 0 - [0, 1/6]∫3x²/√(1-36x²) dx
π/144 - [0, 1/6]∫3x²/√(1-36x²) dx
π/144 - 1/12 [0, 1/6]∫36x²/√(1-36x²) dx

Now, we make the following substitution: θ = arcsin (6x), so sin θ = 6x, and cos θ dθ = 6 dx. Finding the limits of integration, we see that when x=0, θ=0, and when x=1/6, θ = π/2, so we have:

π/144 - 1/72 [0, π/2]∫sin² θ/√(1-sin² θ) cos θ dθ

Now simplifying:

π/144 - 1/72 [0, π/2]∫sin² θ/√cos² θ cos θ dθ
π/144 - 1/144 [0, π/2]∫1 - cos (2θ) dθ
π/144 - 1/144 (θ - sin (2θ)/2) |[0, π/2]
π/144 - (π/288 - 0 - (0-0))
π/288

#2: This one can be done just using integration by parts twice:

[0, 1]∫x² (e^x)^(1/4) dx
[0, 1]∫x² e^(x/4) dx

Let u=x², du = 2x dx, dv = e^(x/4) dx, v = 4e^(x/4):

4x²e^(x/4)|[0, 1] - [0, 1]∫8x e^(x/4) dx
4e^(1/4) - [0, 1]∫8x e^(x/4) dx

Let u=8x, du = 8 dx, dv = e^(x/4) dx, v = 4e^(x/4):

4e^(1/4) - 8xe^(x/4)|[0, 1] + [0, 1]∫32 e^(x/4) dx
4e^(1/4) - 32e^(1/4) + [0, 1]∫32 e^(x/4) dx
-28e^(1/4) + 128 e^(x/4)|[0, 1]
-28e^(1/4) + 128 e^(1/4) - 128
100e^(1/4) - 128

And we are done.

Answer 3

hi bill ok, the two the solutions arrived at are one and an identical ya. the two would desire to be simplified as ? e^x sin x dx = [e^x (sin x - cos x)] / 2 ?sin(x)e^x dx= -(e^x)cos(x) + [(e^x)(sin(x)) - ?sin(x)e^x dx] After simplification, it is going to become as you wrote 2?sin(x)e^x dx= -(e^x)cos(x) + (e^x)(sin(x)) extra simplified into 2?sin(x)e^x dx= (e^x)[(sin(x) - cos x)] So ? e^x sin x dx = [e^x (sin x - cos x)] / 2 surprising. you have written the stairs so needless to say. yet merely made a slip interior the previous couple of steps. Congrats and all the excellent ya.

Answer 4

1♥ my numeric integration gives exactly pi/288;
2♠ both calculus and numeric give 0.4025;
3♣ start voting yourself or no more help for whiz-kids;