<jupyter_start><jupyter_text>Author: <NAME><jupyter_code>import sympy as smp
from sympy import *
from sympy.interactive import printing
printing.init_printing(use_latex = True)
from sympy import Eq, solve_linear_system, Matrix
from numpy import linalg
import numpy as np

x, y, c = smp.symbols('x y c')<jupyter_output><empty_output><jupyter_text>Limits $$\lim_{x \to \pi} \sin(x/2+ \sin(x))$$<jupyter_code>smp.limit(smp.sin(x/2 + smp.sin(x)), x, smp.pi)<jupyter_output><empty_output><jupyter_text>$$ \lim_{x \to 0^+} \frac{2e^{1/x}}{e^{1/x}+1} $$<jupyter_code>smp.limit(2*smp.exp(1/x) / (smp.exp(1/x)+1), x, 0, dir='+')<jupyter_output><empty_output><jupyter_text>$$ \lim_{x \to 0^-} \frac{2e^{1/x}}{e^{1/x}+1} $$<jupyter_code>smp.limit(2*smp.exp(1/x) / (smp.exp(1/x)+1), x, 0, dir='-')<jupyter_output><empty_output><jupyter_text>$$ \lim_{x \to \infty} \frac{\cos(x)-1}{x} $$<jupyter_code>smp.limit((smp.cos(x) - 1)/x, x, smp.oo)<jupyter_output><empty_output><jupyter_text>Derivatives $$ \frac{d}{dx} \left( \frac{1+\sin x}{1 - \cos x} \right)^2 $$<jupyter_code>smp.diff(((1 + smp.sin(x)) / (1 - smp.cos(x)))**2, x)<jupyter_output><empty_output><jupyter_text>$$ \frac{d}{dx} (\log_5 (x))^{x/2}$$<jupyter_code>smp.diff(smp.log(x, 5)**(x/2), x)<jupyter_output><empty_output><jupyter_text>$$\frac{d}{dx} f(x+g(x)) $$<jupyter_code>f, g = smp.symbols('f g', cls=smp.Function)
g = g(x)
f = f(x+g)
smp.diff(f,x)<jupyter_output><empty_output><jupyter_text>Basic Antiderivatives $$ \int \csc(x) \cot (x)  dx $$<jupyter_code>smp.integrate(smp.csc(x)*smp.cot(x), x)<jupyter_output><empty_output><jupyter_text>$$ \int 4 \sec (3x) \tan(3x)dx $$<jupyter_code>smp.integrate(4*smp.sec(3*x)*smp.tan(3*x), x)<jupyter_output><empty_output><jupyter_text>$$ \int \left( \frac{2}{\sqrt{1-x^2}} - \frac{1}{x^{1/4}}\right) dx$$<jupyter_code>smp.integrate(2/smp.sqrt(1-x**2) - 1/x**smp.Rational(1,4), x)<jupyter_output><empty_output><jupyter_text>Initial Value Problems 1. Given $\frac{dy}{dx} = 8x + \csc^2(x)$ with $y(\pi/2) = -7$  solve for $y(x)$<jupyter_code>integral = smp.integrate(8*x + smp.csc(x)**2, x)
C = - integral.subs(x, smp.pi/2) - 7
y = integral + C
y.subs(x, smp.pi/2)
y<jupyter_output><empty_output><jupyter_text>More Complicated Expressions  $$\int \frac{(1+\sqrt{x})^{1/3}}{\sqrt{x}} dx$$<jupyter_code>smp.integrate((1+smp.sqrt(x))**smp.Rational(1,3) / smp.sqrt(x), x)<jupyter_output><empty_output><jupyter_text>$$\int x (1-x^2)^{1/4} dx$$<jupyter_code>smp.integrate(x*(1-x**2)**smp.Rational(1/4), x)<jupyter_output><empty_output><jupyter_text>$$ \int \frac{(2x-1)\cos(\sqrt{3(2x-1)^2+6})}{\sqrt{3(2x-1)^2+6}} dx $$<jupyter_code>smp.integrate((2*x -1)*smp.cos(smp.sqrt(3*(2*x-1)**2 + 6)) / smp.sqrt(3*(2*x-1)**2 + 6), x)<jupyter_output><empty_output><jupyter_text>Definite Integrals $$\int_{0}^{\ln(4)}\frac{ce^x dt}{\sqrt{e^{2x}+9}} $$<jupyter_code>smp.integrate(smp.exp(x) / smp.sqrt(smp.exp(2*x) + 9), (x, 0, smp.log(4)))
l= smp.symbols('l')
a = (x**3 - 3*(L**2)*x)
smp.integrate(smp.exp(x) / smp.sqrt(smp.exp(2*x) + 9), (x, 0, smp.log(4)))<jupyter_output><empty_output><jupyter_text>$$\int_1^t x^{10} e^x dx $$<jupyter_code>l= smp.symbols('l')
C2 = smp.symbols('C2')
C3 = smp.symbols('C3')
m = smp.symbols('m')
W2 = (x**3 - 3*(L**2)*x)
W1 = (x**2 - 2*l*x)
Rd = (2*C2 + 6*C3*x) - m+( 270 + C2*W1 + C3*W2)

smp.integrate( W1*Rd, (x,0,0.001))
smp.integrate( W2*Rd, (x,0,0.001))
smp.integrate(x**10*smp.exp(x), (x, 1, t))<jupyter_output><empty_output><jupyter_text>Improper Integrals $$ \int_{0}^{\infty} \frac{16 \tan^{-1}(x)}{1+x^2} dx $$<jupyter_code>smp.integrate(16*smp.atan(x) / (1+x**2), (x, 0, smp.oo))<jupyter_output><empty_output><jupyter_text>Sequences and Series<jupyter_code>n = smp.symbols('n')<jupyter_output><empty_output><jupyter_text>$$ \sum_{n=0}^\infty \frac{6}{4^n} $$<jupyter_code>smp.Sum(6/4**n, (n,0,smp.oo)).doit()<jupyter_output><empty_output><jupyter_text>$$ \sum_{n=0}^\infty \frac{2^{n+1}}{5^n} $$<jupyter_code>smp.Sum(2**(n+1) / 5**n, (n,0,smp.oo)).doit()<jupyter_output><empty_output><jupyter_text>$$ \sum_{n=1}^{\infty} \frac{\tan^{-1}(n)}{n^{1.1}} $$<jupyter_code>smp.Sum(smp.atan(n)/n**smp.Rational(11,10), (n,1,smp.oo)).doit()<jupyter_output><empty_output><jupyter_text>$$ \sum_{n=1}^\infty \frac{1 + \cos(n)}{n^2}$$<jupyter_code>smp.Sum((1+smp.cos(n))/n**2, (n,1,smp.oo)).n()
m = 400
l= 1

2*((l**3)/3 - l**3)
m*((l**5)/5-3*(l**5)/2 + 4*(l**5)/3)
(6/4)*(l**4) -m*((l**6)/6 -3/4*(l**6)) -2*l*(2*l**3 -m*(-4/5*(l**5)))
((l**3)/3-l**3)*270*m<jupyter_output><empty_output>
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