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I think I am supposed to use integration by substitution or integration by parts, but I'm not sure.

Can anyone help with this?

$\int{ x^3\cdot\sqrt{x^2-1}\ dx }$

short way:

Substitution

$\begin{array}{|rcll|} \hline u &=& \sqrt{x^2-1}=(x^2-1)^{\frac12} \quad & \text{or} \quad x = \sqrt{u^2+1} \\ du &=& \frac12 \cdot (x^2-1)^{-\frac12} \cdot 2x\ dx \\ &=& \frac{x}{\sqrt{x^2-1}}\ dx \quad & \text{or} \quad dx = \frac{\sqrt{x^2-1}}{x}\ du\\ \hline \end{array} \begin{array}{|rcll|} \hline && \int{ x^3\cdot\sqrt{x^2-1}\ dx } \\ &=& \int{ ( \sqrt{u^2+1})^3\cdot u \frac{\sqrt{x^2-1}}{x}\ du } \\ &=& \int{ ( \sqrt{u^2+1})^3\cdot u \frac{u}{\sqrt{u^2+1}}\ du } \\ &=& \int{ ( \sqrt{u^2+1})^2\cdot u^2\ du }\\ &=& \int{ (u^2+1)\cdot u^2\ du }\\ &=& \int{ (u^4+u^2)\ du }\\ &=& \frac{u^3}{3} + \frac{u^5}{5} + c \\ &=& \frac{u^3}{3}\cdot \frac55 + \frac{u^5}{5}\cdot \frac33 + c \\ &=& \frac{u^3}{15}\cdot \left( 5+3u^2 \right)+ c \\ \hline \end{array}$

Back substitution

$\begin{array}{|rcll|} \hline u &=& \sqrt{x^2 -1} \\ \hline \end{array} \begin{array}{|rcll|} \hline && \mathbf{\int{ x^3\cdot\sqrt{x^2-1}\ dx } }\\ &=& \frac{u^3}{15}\cdot \left( 5+3u^2 \right)+ c \\ &=& \frac{(\sqrt{x^2 -1})^3}{15}\cdot \left[ 5+3(\sqrt{x^2 -1})^2 \right]+ c \\ &=& \frac{(x^2 -1)^{\frac32}}{15}\cdot \left[ 5+3(x^2 -1) \right]+ c \\ &=& \frac{(x^2 -1)^{\frac32}}{15}\cdot ( 5+3x^2-3 )+ c \\ &\mathbf{=}& \mathbf{ \dfrac{(x^2 -1)^{\frac32}}{15}\cdot ( 3x^2+2 )+ c }\\ \hline \end{array}$

I think I am supposed to use integration by substitution or integration by parts, but I'm not sure. Can anyone help with this?

$\int{x}^{3}\sqrt{{x}^{2}-1}dx$

We apply twice substitution

We need the following formulae:

$\begin{array}{|rcll|} \hline \cosh^2(z) - \sinh^2(z) &=& 1 \\ \sinh^2(z) &=& \cosh^2(z) -1 \\ \sinh(z) &=& \sqrt{\cosh^2(z) -1} \\\\ \cosh^2(z) - \sinh^2(z) &=& 1 \\ \cosh^2(z) &=& 1+\sinh^2(z) \\\\ \frac{d}{dz}\sinh(z) &=& \cosh(z) \\ \frac{d}{dz}\cosh(z) &=& \sinh(z) \\ \hline \end{array}$

1. Substitution

$\begin{array}{|rcll|} \hline x &=& \cosh(z)\\ dx &=& \sinh(z)\ dz \\ \hline \end{array} \begin{array}{|rcll|} \hline && \int{ x^3\cdot\sqrt{x^2-1}\ dx } \\ &=& \int{ \cosh^3(z)\cdot \sqrt{\cosh^2(z)-1} \cdot \sinh(z)\ dz } \\ &=& \int{ \cosh^3(z)\cdot \sinh(z) \cdot \sinh(z)\ dz } \\ &=& \int{ \cosh^3(z)\cdot \sinh^2(z)\ dz } \\ \hline \end{array}$

2. Substitution

$\begin{array}{|rcll|} \hline u &=& \sinh(z)\\ du &=& \cosh(z)\ dz \\ \hline \end{array} \begin{array}{|rcll|} \hline && \int{ \cosh^3(z)\cdot \sinh^2(z)\ dz } \\ &=& \int{ \cosh^2(z)\cdot \sinh^2(z)\cdot \cosh(z)\ dz \\ } \\ &=& \int{ [1+\sinh^2(z)]\cdot \sinh^2(z)\cdot \cosh(z)\ dz \\ } \\ &=& \int{ (1+u^2)\cdot u^2 \ du } \\ &=& \int{ (u^2+u^4)\ du } \\ &=& \frac{u^3}{3} + \frac{u^5}{5} + c \\ &=& \frac{u^3}{3} + \frac{u^5}{5} + c \\ &=& \frac{u^3}{3}\cdot \frac55 + \frac{u^5}{5}\cdot \frac33 + c \\ &=& \frac{u^3}{15}\cdot \left( 5+3u^2 \right)+ c \\ \hline \end{array}$

3. Back substitution

$\begin{array}{|rcll|} \hline u &=& \sinh(z) \\ &=& \sqrt{\cosh^2(z) -1} \\ &=& \sqrt{x^2 -1} \\ \hline \end{array} \begin{array}{|rcll|} \hline && \mathbf{\int{ x^3\cdot\sqrt{x^2-1}\ dx } }\\ &=& \frac{u^3}{15}\cdot \left( 5+3u^2 \right)+ c \\ &=& \frac{(\sqrt{x^2 -1})^3}{15}\cdot \left[ 5+3(\sqrt{x^2 -1})^2 \right]+ c \\ &=& \frac{(x^2 -1)^{\frac32}}{15}\cdot \left[ 5+3(x^2 -1) \right]+ c \\ &=& \frac{(x^2 -1)^{\frac32}}{15}\cdot ( 5+3x^2-3 )+ c \\ &\mathbf{=}& \mathbf{ \dfrac{(x^2 -1)^{\frac32}}{15}\cdot ( 3x^2+2 )+ c }\\ \hline \end{array}$

How does sqrt(x-3)^2=x-3?

1.
$\begin{array}{|rcll|} \hline \sqrt{(x-3)^2} &=& \left(~ \sqrt{x-3} ~ \right)^2 \\ \hline \end{array}$

2.

$\begin{array}{|rcll|} \hline \sqrt{x-3} &=& (x-3)^{\frac12} \\ \hline \end{array}$

3.

$\begin{array}{|rcll|} \hline \sqrt{(x-3)^2} &=& \left( ~ \sqrt{x-3} ~ \right)^2 \\ &=& \left( ~ (x-3)^{\frac12} ~ \right)^2 \\ &=& (x-3)^{\frac12\cdot 2} \\ &=& (x-3)^{\frac22 } \\ &=& (x-3)^{1} \\ &=& (x-3) \\ \hline \end{array}$

what is the answer of this question: a2 +b2 = c2 14 square + 14 square 196 + 196 = 392

Then find the square root of 392

$\sqrt{392} = 19.7989898732$

S₁₂ for the series 1 + 1.5 + 2.25 + 3.375 +

$\begin{array}{|rcll|} \hline && 1 + 1.5 + 2.25 + 3.375 +\ldots \\ &=& 1 + 1.5 + 1.5^2 + 1.5^3 + 1.5^4 + 1.5^5 + \ldots \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline A = s_{12} = &=& 1 + 1.5 + 1.5^2 + 1.5^3 + 1.5^4 + 1.5^5 +\ldots + 1.5^{10} + 1.5^{11} \\ B = 1.5\cdot s_{12} = &=& 1.5 + 1.5^2 + 1.5^3 + 1.5^4 + 1.5^5 + 1.5^6 +\ldots + 1.5^{11} + 1.5^{12} \\ B-A = 1.5\cdot s_{12}-s_{12} = &=& 1.5^{12} -1 \\ s_{12}\cdot (1.5-1)&=& 1.5^{12} -1 \\ s_{12}\cdot 0.5&=& 1.5^{12} -1 \\ s_{12} &=& \frac{1.5^{12} -1}{0.5} \\\\ s_{12} &=& \frac{128.746337891}{0.5} \\\\ \mathbf{ s_{12}} &\mathbf{=}& \mathbf{257.492675781\ldots} \\ \hline \end{array}$

Given: 

sin a= -21/29 , with 3pie/2 <a <2pie

and

tan B= -24/7 ,with pie/2 <B< pie

Find cos(a+b)

1.

$\begin{array}{|rcll|} \hline \sin(a) &=& \frac{-21}{29}, \text{ with } \frac{3\pi}{2} <a< 2\pi \quad \text{IV. Quadrant} \\ a &=& \arcsin(\frac{-21}{29}) \\ \mathbf{a} & \mathbf{=} & \mathbf{-0.80978357257\ \text{rad } \quad (-46.3971810273^{\circ}) }\\ \hline \end{array}$

2.

$\begin{array}{|rcll|} \hline \tan(b) &=& \frac{24}{-7}, \text{ with } \frac{\pi}{2} <b< \pi \quad \text{II. Quadrant} \\ b &=& \arctan(\frac{24}{-7}) \\ b &=& -1.28700221759 + \pi \quad (\text{II. Quadrant}) \\ \mathbf{b} & \mathbf{=} & \mathbf{1.85459043600 \ \text{rad } \quad (106.260204708^{\circ}) }\\ \hline \end{array}$

3.

$\begin{array}{|rcll|} \hline \cos(a+b) &=& \cos(-0.80978357257\ \text{rad}+1.85459043600 \ \text{rad } ) \\ &=& \cos(1.04480686343 \ \text{rad } ) \\ \mathbf{\cos(a+b)} & \mathbf{=} & \mathbf{0.50206896552} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && 30\ \text{cm} \cdot \dfrac{13€}{10000\ \text{cm}} \\\\ &=& 30 \cdot \dfrac{13}{10000} € \\\\ &=& \dfrac{30 \cdot 13}{10000} € \\\\ &=& \dfrac{390}{10000} € \\\\ &=& 0,039 € \\\\ &=& 3,9\ \text{cent} \\ \hline \end{array}$

3/(x-2)-1/(x+2)=5

$\begin{array}{|rcll|} \hline 5 &=& \frac{3}{x-2}-\frac{1}{x+2} \quad & | \quad \cdot (x+2)(x-2) \\ 5\cdot(x+2)(x-2) &=& \frac{ 3(x+2)(x-2)}{x-2}-\frac{(x+2)(x-2) }{x+2} \\ 5\cdot (x+2)(x-2) &=& 3(x+2) - (x-2) \\ 5\cdot (x+2)(x-2)) &=& 3x+ 6 - x+2 \\ 5\cdot (x+2)(x-2) &=& 2x+8 \\ 5\cdot (x^2-4) &=& 2x+8 \\ 5x^2-20 &=& 2x+8 \quad & | \quad -2x-8 \\ 5x^2-20 -2x-8 &=& 0 \\ 5x^2-2x -28 &=& 0 \\ \\ \begin{array}{|rcll|} \hline ax^2+bx+c &=& 0 \\ x &=& \frac{-b\pm\sqrt{b^2-4ac} }{2a} \\ \hline \end{array} \\ ax^2+bx+c &=& 0 \quad & | \quad a = 5 \quad b = -2 \quad c=-28 \\ x &=& \frac{ 2\pm\sqrt{4-4\cdot 5 (-28)} }{2\cdot 5} \\ x &=& \frac{2\pm\sqrt{4+20\cdot28)} }{10} \\ x &=& \frac{2\pm\sqrt{564} }{10} \\ x &=& \frac{2\pm 23.7486841741 }{10} \\\\ x_1 & = & \frac{2+ 23.7486841741 }{10} \\ \mathbf{x_1} & \mathbf{=} & \mathbf{2.57486841741 } \\\\ x_2 &=& \frac{2- 23.7486841741 }{10} \\ \mathbf{x_2} & \mathbf{=} & \mathbf{-2.17486841741} \\ \hline \end{array}$

2. If θ is a first-quadrant angle in standard position with P(u, v) = (3, 4), evaluate sin 2 θ.

3. If θ is a first-quadrant angle in standard position with P(u, v) = (3, 4), evaluate cos 2 θ.

$\begin{array}{|rcll|} \hline r^2 &=& (4-r)^2+3^2 \\ r^2 &=& 4^2-8r+r^2+3^2 \\ 8r &=& 4^2+3^2 \\ 8r &=& 5^2 \\ r &=& \frac{5^2}{8} \\ r &=& \frac{25}{8} \\ \hline \end{array}$

2. If θ is a first-quadrant angle in standard position with P(u, v) = (3, 4), evaluate sin 2 θ.

$\begin{array}{|rcll|} \hline \sin(2\theta) &=& \frac{3}{r} \\ \sin(2\theta) &=& \frac{3}{\frac{25}{8}} \\ \sin(2\theta) &=& \frac{24}{25} \\ \hline \end{array}$

3. If θ is a first-quadrant angle in standard position with P(u, v) = (3, 4), evaluate cos 2 θ.

$\begin{array}{|rcll|} \hline \cos(2\theta) &=& \frac{4-r}{r} \\ \cos(2\theta) &=& \frac{4}{r}-1 \\ \cos(2\theta) &=& \frac{4\cdot 8}{25}-1 \\ \cos(2\theta) &=& \frac{32}{25}-1 \\ \cos(2\theta) &=& \frac{7}{25} \\ \hline \end{array}$

How to find the inverse of a cubic function?

$\begin{array}{|lrcll|} \hline 1. & y &=& 2(x-3)^3-16 \\\\ \text{Inverse:} & x &=& 2(y-3)^3-16 \quad & | \quad +16 \\ & x+16 &=& 2(y-3)^3 \quad & | \quad :2 \\ & \frac{x +16}{2}&=& (y-3)^3 \\ & \frac{x}{2} + 8 &=& (y-3)^3 \quad & | \quad \text{cube root both sides} \\ & \sqrt[3]{\frac{x}{2} + 8} &=& y-3 \quad & | \quad +3 \\ & \sqrt[3]{\frac{x}{2} + 8}+3 &=& y \\ & y &=& \sqrt[3]{\frac{x}{2} + 8}+3 \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline 2. & f(x) &=& \sqrt[3]{x + 1}+5 \\\\ \text{Inverse:} & x &=& \sqrt[3]{y + 1}+5 \quad & | \quad -5 \\ & x-5 &=& \sqrt[3]{y + 1} \quad & | \quad \text{cube both sides} \\ & (x-5)^3 &=& y + 1 \quad & | \quad -1 \\ & (x-5)^3-1 &=& y \\ & y &=& (x-5)^3-1 \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline 3. & y &=& \frac12(x-1)^3+3 \\\\ \text{Inverse:} & x &=& \frac12(y-1)^3+3 \quad & | \quad -3 \\ & x-3 &=& \frac12(y-1)^3 \quad & | \quad \cdot 2 \\ & 2(x-3) &=& (y-1)^3 \\ & 2(x-3) &=& (y-1)^3 \quad & | \quad \text{cube root both sides} \\ & \sqrt[3]{2(x-3)} &=& y-1 \quad & | \quad +1 \\ & \sqrt[3]{2(x-3)}+1 &=& y \\ & y &=& \sqrt[3]{2(x-3)}+1 \\ \hline \end{array}$