01 - Standard Integration Techniques

Integration by Substitution

Let $g:[a,b]\to I$ be continuously differentiable and $f:I\to \mathbb{R}$ be continuous
Then

$${\int }_a^{b}f(g(x))g^{\prime }(x)dx={\int }_{g(a)}^{g(b)}f(u)du$$

where $u=g(x)$

Example (better to see how to use)

Evaluate

$$I={\int }_0^1\frac{1}{\sqrt{4-2x-x^2}}dx$$

Answer

As the integrand is very similar to

$$\frac{1}{\sqrt{1-x^2}}$$

whose integral we know as to be ${\sin }^{-1}x+c$

Then we write

$$\begin{array}{rl}4-2x-x^2& =5-(x+1{)}^2\\ & =5\left(1-{\left(\frac{x+1}{\sqrt{5}}\right)}^2\right)\end{array}$$

We can use the substitution $t=\frac{x+1}{\sqrt{5}}$ then

$$\begin{array}{rl}I& ={\int }_0^1\frac{1}{\sqrt{4-2x-x^2}}dx\\ & =\frac{1}{\sqrt{5}}{\int }_0^1\frac{1}{\sqrt{1-{\left(\frac{x+1}{\sqrt{5}}\right)}^2}}dx\\ & =\frac{1}{\sqrt{5}}{\int }_{\frac{1}{\sqrt{5}}}^{\frac{2}{\sqrt{5}}}\frac{\sqrt{5}}{\sqrt{1-t^2}}dt\\ & ={\int }_{\frac{1}{\sqrt{5}}}^{\frac{2}{\sqrt{5}}}\frac{1}{\sqrt{1-t^2}}dt\\ & ={\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)-{\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)\end{array}$$

Integration by Parts

Note that it is simply the integral form of the product rule

$${\int }_a^{b}f(x)g^{\prime }(x)dx=[f(x)g(x){]}_a^b-{\int }_a^{b}g(x)f^{\prime }(x)dx$$

Reduction Formulae (part of Integration by Parts)

Consider

$$I_n=\int {\cos }^{n}xdx$$

where $n$ is non-negative

Find a reduction formula for $I_n$

Proof

We aim to write $I_n$ in terms of other $I_k$ for $k<n$
By Integration by Parts then

$$\begin{array}{rl}{I}_n& =\int {\cos }^{n-1}x\cdot \cos xdx\\ & ={\cos }^{n-1}x\cdot \sin x-\int (n-1){\cos }^{n-2}x\cdot (-\sin x)\sin xdx\\ & ={\cos }^{n-1}x\cdot \sin x+(n-1)\int {\cos }^{n-2}x(1-{\cos }^{2}x)\\ & ={\cos }^{n-1}x\cdot \sin x+(n-1)(I_{n-2}-I_n)\end{array}$$

Hence

$$I_n=\frac{1}{n}{\cos }^{n-1}x\cdot \sin x+\frac{n-1}{n}{I}_{n-2}$$

for $n\ge 2$