Why is ln x leq sqrt x

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Calculate $\ D \ int_0 ^ \ infty \ frac {1} {\ sqrt {1 + x ^ 2} ^ 3} \, dx$ by first using the substitution $x = \ sinh t$ and then $u = substitute e ^ t$.
tip

$\ D 1+ \ sinh ^ 2 t = \ cosh ^ 2 \ \ D \ cosh t = \ frac 12 \ left (e ^ t + \ frac {1} {e ^ t} \ right)$

solution

$\ D x = \ sinh t, \ quad \ frac {dx} {dt} = \ cosh t = \ sqrt {1 + x ^ 2}, \ quad x = 0 \ LR t = 0, \ quad x \ to \ infty \ LR t \ to \ infty \ \ D \ RA \ int_0 ^ \ infty \ frac {1} {\ sqrt {1 + x ^ 2} ^ 3} \, dx = \ int_0 ^ \ infty \ frac { 1} {\ cosh ^ 3 t} \ cosh t \, dt = \ int_0 ^ \ infty \ frac {1} {\ cosh ^ 2 t} \, dt \ \ D e ^ t = u, \ quad \ frac {du} {dt} = e ^ t = u, \ quad t = 0 \ LR u = 1, \ quad t \ to \ infty \ LR u \ to \ infty \ \ D \ RA \ int_0 ^ \ infty \ frac {1} {\ cosh ^ 2 t} \, dt = \ int_1 ^ \ infty \ frac {4} {\ left (u + \ frac 1u \ right) ^ 2} \ frac 1u \, du = \ int_1 ^ \ infty \ frac {4u} {\ left (u ^ 2 + 1 \ right) ^ 2} \, du \ \ D w = u ^ 2 + 1, \ quad \ frac {dw} {du} = 2u, \ quad u = 1 \ LR w = 2, \ quad u \ to \ infty \ LR w \ to \ infty \ \ D \ RA \ int_1 ^ \ infty \ frac {4u} {\ left (u ^ 2 + 1 \ right) ^ 2} \, du = 2 \ int_2 ^ \ infty \ frac {1} {w ^ 2} \, dw = 2 \ left [- \ frac 1w \ right] _2 ^ \ infty = 1$

exercise 2

Find an antiderivative to $\ D f (x) = \ left (\ frac {x} {\ sqrt {1-x ^ 2}} \ right) ^ 3$.
1. with the substitution $u = \ sqrt {1-x ^ 2}$.

2. with the substitutions $x = \ sin t$ and $u = \ cos t$.

tip
1. Case distinction.
2. $\ sin ^ 2 t + \ cos ^ 2 t = 1$.

solution
1. First let $x \ ge 0$.

$\ D u = \ sqrt {1-x ^ 2} \ quad \ LR \ quad x = \ sqrt {1-u ^ 2} \ quad \ mbox {and} \ quad \ frac {dx} {du} = - \ frac {u} {\ sqrt {1-u ^ 2}} \ \ D \ RA \ quad \ int \ frac {x ^ 3} {\ sqrt {1-x ^ 2} ^ 3} \, dx = - \ int \ frac {\ sqrt {1-u ^ 2} ^ 3} {u ^ 3} \, \ frac {u} {\ sqrt {1-u ^ 2}} \, du \ \ D = - \ int \ frac {1-u ^ 2} {u ^ 2} \, du = \ frac {1} {u} + u + C = \ frac {1} {\ sqrt {1-x ^ 2}} + \ sqrt {1-x ^ 2} + C = \ frac {2-x ^ 2} {\ sqrt {1-x ^ 2}} + C$

If $x \ le 0$, then $\ D x = - \ sqrt {1-u ^ 2}$ and $\ D \ frac {dx} {du} = \ frac {u} {\ sqrt {1- u ^ 2}}$. These two changed signs cancel each other out in the second integral,
and the antiderivative has the same form as above.

2. Since $f$ is only defined on $[- 1,1]$, $x = \ sin t$ for $t \ in [- \ frac \ pi 2, \ frac \ pi 2]$ is a bijective mapping. The cosine is positive on this interval.
Therefore one has

$\ D x = \ sin t \ quad \ RA \ quad \ sqrt {1-x ^ 2} = \ cos t \ quad \ mbox {and} \ quad \ frac {dx} {dt} = \ cos t \ \ D \ int \ frac {x ^ 3} {\ sqrt {1-x ^ 2} ^ 3} \, dx = - \ int \ frac {\ sin ^ 3 t} {\ cos ^ 3 t} \ cos t \, dt = \ int \ frac {\ sin ^ 3 t} {\ cos ^ 2 t} \, dt = \ int \ frac {1- \ cos ^ 2 t} {\ cos ^ 2 t} \ sin t \ dt$

With $u = \ cos t$ and $\ D \ frac {du} {dt} = - \ sin t$ the integral becomes

$\ D \ int \ Big (1- \ frac {1} {u ^ 2} \ Big) \, du = u + \ frac {1} {u} + C \ \ D = \ cos (t) + \ frac {1} {\ cos (t)} + C = \ sqrt {1-x ^ 2} + \ frac {1} {\ sqrt {1-x ^ 2}} + C = \ frac {2-x ^ 2} {\ sqrt {1-x ^ 2}} + C.$

Calculate $\ D \ int_0 ^ 1 \ sqrt {1+ \ sqrt {x}} \, dx$
tip

solution

$\ D t = 1 + \ sqrt {x} \ LR x = (t-1) ^ 2, \ quad \ frac {dx} {dt} = 2 (t-1), \ quad x = 0 \ LR t = 1, \ quad x = 1 \ LR t = 2 \ \ D \ RA \ int_0 ^ 1 \ sqrt {1+ \ sqrt {x}} \, dx = \ int_1 ^ 2 \ sqrt {t} \ cdot 2 (t-1) \, dt = 2 \ int \ left (t ^ {\ frac {3} {2}} - t ^ {\ frac {1} {2}} \ right) \, dt \ \ D = 2 \ left [\ frac {2} {5} t ^ {\ frac {5} {2}} - \ frac {2} {3} t ^ {\ frac {3} {2}} \ right] ^ 2_1 \ \ D = 4 \ left [\ frac {2} {5} 2 ^ {\ frac {5} {2}} - \ frac {2} {5} - \ frac {2} {3} 2 ^ {\ frac {3} {2}} + \ frac {2} {3} \ right] = 4 \ left [\ sqrt {2} \ left (\ frac {1} {5} \ cdot4- \ frac {1 } {3} \ cdot2 \ right) + \ left (\ frac {1} {3} - \ frac {1} {5} \ right) \ right] = \ frac {8} {15} (\ sqrt {2 } +1)$

1. $\ D \ int \ frac {\ cos x} {(\ sin x + 1) \ sin x} \, dx \ quad$ Note: Substitute appropriately.

2. $\ D \ int x ^ 2 \ ln x \, dx$

3. $\ D \ int_1 ^ 3 \ frac {x ^ 2-8} {x ^ 2-16} \, dx$

tip
1. Partial fraction decomposition.
2. Partial integration.
3. sent $0 to add$ and PBZ.

solution
1. $\ D t = \ sin x, \ quad \ frac {dt} {dx} = \ cos t \ LR dx = \ frac {1} {\ cos x} dt \ \ D = \ int \ frac {\ cos x} {(t + 1) t} \ frac {1} {\ cos x} \, dt = \ int \ frac {1} {(t + 1) t} \, dt \$

Partial fraction decomposition:

$\ D \ frac {1} {(t + 1) t} = \ frac At + \ frac {B} {t + 1} = \ frac {A (t + 1) + Bt} {t (t + 1) } \ \ D \ RA A (t + 1) + Bt = 1 \ LR A = 1, \ quad B + A = 0 \ LR B = -1 \ \ D = \ int \ frac 1t- \ frac { 1} {t + 1} \, dt = \ ln | t | - \ ln | t + 1 | + c \ D = \ ln | \ sin x | - \ ln | \ sin x + 1 | + c = \ ln \ left (\ frac {| sin x |} {sin x +1} \ right) + c$

2. $\ D \ int x ^ 2 \ ln x \, dx = \ frac 13 x ^ 3 \ ln x- \ int \ frac 13 x ^ 3 \ frac 1x \, dx \ \ D = \ frac 13 x ^ 3 \ ln x- \ frac 13 \ int x ^ 2 \, dx = \ frac 13 x ^ 3 \ ln x- \ frac 19 x ^ 3 + c \ \ D = \ frac 19 x ^ 3 (3 \ ln x -1) + c$

3. $\ D \ int_1 ^ 3 \ frac {x ^ 2-8} {x ^ 2-16} \, dx = \ int_1 ^ 3 \ frac {x ^ 2-16 + 8} {x ^ 2-16} \ , dx = \ int_1 ^ 3 1+ \ frac {8} {x ^ 2-16} \, dx$

Patial fracture dissection

$\ D \ frac {8} {x ^ 2-16} = \ frac {A} {x-4} + \ frac {B} {x + 4} = \ frac {A (x + 4) + B ( x-4)} {x ^ 2 + 16} \ \ D \ RA A + B = 0, \ quad 4A-4B = 8 \ LR A = 1, B = -1 \ \ D = \ int_1 ^ 3 \ left (1+ \ frac {1} {x-4} - \ frac {1} {x + 4} \ right) \, dx = \ big [x + \ ln | x-4 | - \ ln | x + 4 | \ big] _1 ^ 3 = 3 + 0- \ ln 7-1- \ ln 3+ \ ln 5 = 2 + \ ln \ frac {5} {21}$

Calculate
1. $\ D \ int \ frac {\ sin x} {1+ \ cos ^ 2 x} \, dx$

2. $\ D \ int \ frac {6x ^ 3 + 4x ^ 2-4x-16} {x ^ 4-3x ^ 2-4} \, dx$.
The use of an integral board is not permitted for this task.
tip
1. Skilfully substitute.
2. Partial fraction decomposition.

solution
1. $\ D t = \ cos x, \ quad \ frac {dt} {dx} = - \ sin x \ \ D \ RA \ int \ frac {\ sin x} {1+ \ cos ^ 2 x} \, dx = \ int \ frac {\ sin x} {1 + t ^ 2} \ cdot \ left (- \ frac {1} {\ sin x} \ right) \, dt \ \ D = - \ int \ frac {1} {1 + t ^ 2} \, dt = - \ arctan (t) + C = - \ arctan {(\ cos x)} + C$
2. Partial fraction decomposition:

$\ D x ^ 4-3x ^ 2-4 = (x ^ 2 + 1) (x ^ 2-4) = (x ^ 2 + 1) (x-2) (x + 2) \ \ D \ RA f (x) = \ frac {6x ^ 3 + 4x ^ 2-4x-16} {x ^ 4-3x ^ 2-4} = \ frac {Ax + B} {x ^ 2 + 1} + \ frac {C} {x-2} + \ frac {D} {x + 2} \ \ D \ LR 6x ^ 3 + 4x ^ 2-4x-16 = (Ax + B) (x ^ 2-4) + C (x ^ 2 + 1) (x + 2) + D (x ^ 2 + 1) (x-2)$

Substituting $x = 2$ and $x = -2$:

$\ D \ quad 48 + 16-8-16 = \ quad 20C \ LR C = 2 \ \ D-48 + 16 + 8-16 = -20D \ LR D = 2$

Compare the terms with $x ^ 3$ and $x ^ 2$:

$\ D 6x ^ 3 = x ^ 3 \ left (A + 2 + 2 \ right) \ LR A = 2 \ \ D 4x ^ 2 = x ^ 2 \ left (B + 4-4 \ right) \ LR B = 4 \ \ D \ RA f (x) = \ frac {2x + 4} {x ^ 2 + 1} + \ frac {2} {x-2} + \ frac {2} {x + 2} \ \ D \ int f (x) \, dx = 2 \ left (\ ln \ left | x-2 \ right | + \ ln \ left | x + 2 \ right | \ right) + \ int \ frac { 2x} {x ^ 2 + 1} + \ frac {4} {x ^ 2 + 1} \, dx \ \ D = 2 \ ln \ left | x ^ 2-4 \ right | +4 \ arctan x + \ int \ frac {2x} {x ^ 2 + 1} \, dx \ \ D \ quad u = x ^ 2, \ quad \ frac {du} {dx} = 2x \ \ D = 2 \ ln \ left | x ^ 2-4 \ right | +4 \ arctan x + \ int \ frac {1} {u + 1} \, du \ \ D = 2 \ ln \ left | x ^ 2-4 \ right | +4 \ arctan x + \ ln \ left | u + 1 \ right | + c \ \ D = 2 \ ln \ left | x ^ 2-4 \ right | +4 \ arctan x + \ ln \ left (x ^ 2 + 1 \ right) + c$

Exercise 6

Let $f \ in C ^ 1 [0,2 \ pi]$ and $\ D b_k = \ int_0 ^ {2 \ pi} f (x) \ cos kx \, dx$.
Show $\ D \ lim_ {k \ to \ infty} b_k = 0$ using partial integration.
tip

$\ D \ int_a ^ b u '(x) \ cdot v (x) \, dx = u (x) \ cdot v (x) \ big | _a ^ b- \ int_a ^ bu (x) \ cdot v' (x) \, dx$

solution
$\ D b_k = f (x) \ sin (kx) \ frac 1k \ Bigg | _0 ^ {2 \ pi} - \ int_0 ^ {2 \ pi} f '(x) \ sin (kx) \ frac 1k \ , dx \ \ D = \ frac 1k \ left (f (2 \ pi) \ sin (2 \ pi k) -f (0) \ sin (0) \ right) - \ frac 1k \ int_0 ^ {2 \ pi} f '(x) \ sin (kx) \, dx = - \ frac 1k \ int_0 ^ {2 \ pi} f' (x) \ sin (kx) \, dx$

The integral converges because the integrand is continuous and the interval compact, so $b_k$ tends to zero for $k \ to \ infty$ because of the factor $\ frac 1k$.

Exercise 7

Examine $\ D \ int_0 ^ \ infty \ frac {1} {\ sqrt {1 + e ^ x}} \, dx$ for convergence.
Note: it is not necessary to calculate an antiderivative.
tip

Majorant criterion.

solution
$\ D 0 \ leq \ frac {1} {\ sqrt {1 + e ^ x}} \ leq \ frac {1} {\ sqrt {e ^ x}} = e ^ {- \ frac x2} \ \ D \ int_0 ^ \ infty e ^ {- \ frac x2} \, dx = -2e ^ {- \ frac x2} \ Big | _0 ^ \ infty = 2 \ \ RA 0 \ leq \ int_0 ^ \ infty \ frac {1} {\ sqrt {1 + e ^ x}} \, dx \ leq 2$

The integral is not negative and is bounded upwards by the majorant $2$, i.e. it is convergent.

Exercise 8

Let $\ D G (a) = \ int_1 ^ a \ frac {\ ln (1 + a ^ 2t ^ 2)} {t ^ 2} \, dt$. Calculate $G '(a)$. What is $G '(1)$?
tip

$\ D \ int \ frac {1} {1 + x ^ 2} \, dx = \ arctan x + c$

solution

$\ D G '(a) = \ frac {\ ln (1 + a ^ 4)} {a ^ 2} + \ int_1 ^ a \ frac {1} {t ^ 2} \ frac {1} {1+ a ^ 2 t ^ 2} 2at ^ 2 \, dt \ \ D = \ frac {\ ln (1 + a ^ 4)} {a ^ 2} + \ int_1 ^ a \ frac {2a} {1 + a ^ 2 t ^ 2} \, dt \ \ D s = at \ LR t = \ frac sa \ RA t = 1 \ LR s = a, \ quad t = a \ LR s = a ^ 2, \ quad \ frac {dt} {ds} = \ frac 1a \ \ D = \ frac {\ ln (1 + a ^ 4)} {a ^ 2} + \ int_a ^ {a ^ 2} \ frac {2a} {1 + s ^ 2} \ frac 1a \, ds \ \ D = \ frac {\ ln (1 + a ^ 4)} {a ^ 2} + \ int_a ^ {a ^ 2} \ frac {1} {1 + s ^ 2} \, ds \ \ D = \ frac {\ ln (1 + a ^ 4)} {a ^ 2} +2 (\ arctan a ^ 2- \ arctan a) \ \ D G ' (1) = \ ln 2$

Exercise 9

Examine the following improper integrals for convergence:
1. $\ D \ int_1 ^ \ infty e ^ {\ frac {1} {x}} \, dx$
2. $\ D \ int _ {- 1} ^ 0 e ^ {\ frac {1} {x}} \, dx$
3. $\ D \ int_0 ^ \ infty \ frac {\ sin x} {1 + x ^ 3} \, dx$

tip

Find integrable majorante or minorante.

solution
1. The integrand is monotonically decreasing from $e ^ 1$ to $e ^ 0 = 1$, therefore $e ^ \ frac {1} {x} \ ge 1$ and the integral does not exist.
2. It is $\ D \ lim_ {x \ to 0-} e ^ \ frac {1} {x} = 0$. Therefore the integrand $f (x) = e ^ \ frac {1} {x}$ can be continued continuously through $f (0) = 0$ and the integral exists.
Alternative:

$\ D t = \ frac {1} {x}, \ quad x = \ frac {1} {t}, \ quad \ frac {dx} {dt} = - \ frac {1} {t ^ 2} \ \ \ D \ int _ {- 1} ^ 0 e ^ \ frac {1} {x} \, dx = - \ int _ {- 1} ^ {- \ infty} \ frac {1} {t ^ 2} e ^ t \, dt = \ int _ {- \ infty} ^ {- 1} \ frac {1} {t ^ 2} e ^ t \, dt.$

Now $e ^ t$ (or $\ frac {1} {t ^ 2}$) is an integrable majorante.

3. It is

$\ D \ int_0 ^ \ infty \ frac {\ sin x} {1 + x ^ 3} \, dx = \ int_0 ^ 1 \ frac {\ sin x} {1 + x ^ 3} \, dx + \ int_1 ^ \ infty \ frac {\ sin x} {1 + x ^ 3} \, dx$.

The first integral exists because continuous functions can be integrated on compact intervals, in the second $\ frac {1} {x ^ 3}$ is an integrable majorant.

Exercise 10

Let $\ D f (x) = \ frac {1} {1 + x ^ 2}$. We are looking for an approximation for $\ D \ int_0 ^ 1f (x) \, dx$.
Only the function values ​​at the positions $x = 0$, $x = \ frac 12$ and $x = 1$ should be used.

1. What is the value of the summed trapezoidal rule?

2. Prove that the error is less than $\ frac {1} {16}$.

3. What is the value of Kepler's barrel rule?

4. What is the exact value of the integral?

tip
1. Summed trapezoidal rule:

$\ DT ^ {(n)} (f) = h \ left (\ frac 12 f (a) + \ frac 12 f (b) + \ sum_ {i = 1} ^ {n-1} f (a + ih) \ right), \ quad h = \ frac {ba} {n}$
2. Remaining term of the summed trapezoidal rule:

$\ DE ^ {(n)} (f) = - \ frac {(ba)} {12} h ^ 2 f '' (\ xi), \ quad \ text {with suitable intermediate point} \ xi \ in [a , b]$
3. Kepler's barrel rule (Simpson rule):

$\ D S (f) = \ frac {b-a} {6} \ left (f (a) + 4f (\ frac {a + b} {2} + f (b) \ right).$