AntiDerivative

• An indefinite Integral

Integration Tips

• Convert the terms into a sum of more easily integrable terms
  • Then integrate each term, which you can do because of Integral#Properties
• Convert Trigonometry functions into the form $\cos (a+bx)$
  • Works for even powers of trig functions (use cos double angle identity)
• Perform Partial Fractions decomposition if you have a fraction with top degree bigger than bottom
  • Helps because $\pm$ can split up an integral
• Graph
• Split up absolute values by roots, then evaluate the integrals separately
• Remember to Absolute Value the inside of $\ln$

U-Substitution

• We basically hack the chain rule
• So set $u=g(x),\frac{\mathrm{d}u}{\mathrm{d}x}=g^{\prime }(x),\mathrm{d}u=g^{\prime }(x)\mathrm{d}x$
• Replace something of the form $\int f^{\prime }(g(x))g^{\prime }(x)dx$
• Substituting $u$ gives $\int f^{\prime }(u)du=f(u)+C=f(g(x))+C$
• You may need to take a constant outside the integral
• For a definite integral, make sure to update the limits to $g(a),g(b)$
• sometimes, if you get the expression for $u$ in the derivative, it's helpful to sub in $u$
• When you do a substitution, you have to update the bounds of the integral based on $dx=g^{\prime }(u)du$

Double Integrals

• For double integrals, we want to do a one-to-one mapping from $(x,y)$ to $(u,v)$
• When doing this with double integrals, it can also help make the region easier
  • Especially when you can get $(u,v)$ to be some constant
  • Make sure to convert all boundaries in terms of $u$ and $v$
• $dxdy=\left|\frac{\partial (x,y)}{\partial (u,v)}\right|dudv$
• Where $\frac{\partial (x,y)}{\partial (u,v)}$ is the Jacobian
• As long as this doesn't ever go to $0$ strictly within the region (it's fine if it's 0 at the boundary), then we can define
• $$$$

\begin{align}
\frac{ \mathrm{d}(uv) }{ \mathrm{d}x } & =u\frac{ \mathrm{d}v }{ \mathrm{d}x }+v\frac{ \mathrm{d}u }{ \mathrm{d}x } \
\int u\frac{ \mathrm{d}v }{ \mathrm{d}x } +v\frac{ \mathrm{d}u }{ \mathrm{d}x } , dx & =uv \
\boxed{\int u , dv =uv-\int v , du }
\end