Integration technique often used for integrating:

• Inverse Function
• Product Functions

Integration By Parts

Differentiate a function that is the product of 2 simpler functions. It is an Iterative Process $\int u{v}^{\prime }=uv-\int u^{\prime }v$ or $\int udv=uv-\int vdu$ Where:

• $u=f(x)$ is Differentiable
• $v=g(x)$ is Differentiable

Solving Process

1. Pick $u$ and $dv$
   1. Usually pick $u$ as a Logarithms or Inverse Trigonometric Functions if there is one in the LHS
2. Compute $du$ and $v$
3. Check $\int udv=uv-\int vdu$
4. Remember there is one $+c$ at the very end

Tricks

Since Integration by parts is very guess-heavy, you tend to be able to see if you are on the right track early on by your computations of $v,du$

Wrong Example

Determine $\int x\sin xdx$ 5. $u^{\prime }=x$, and then naturally, $v=\sin x$ 6. Therefore, $u=\frac{x^2}{2}+c$ as the antiderivative of $u^{\prime }$. $v^{\prime }=\cos x$ 7. Check $uv-\int u{v}^{\prime }dx=\frac{x^2}{2}\sin x-\int \frac{x^2}{2}dx$. 1. It is too hard to integrate $\int \frac{x^2}{2}dx$, therefore, this was a bad decision of $u^{\prime }$ and $v$

Right Example

Determine $\int x\sin xdx$ 8. $u^{\prime }=\sin x$, $v=x$ 9. $u=-\cos x$, $v^{\prime }=1$ 10. $uv-\int u{v}^{\prime }dx=-x\cos x-\int (-\cos x)dx$ 11. Now solve, $\int x\sin xdx=-x\cos x+\sin x+c$