New Notes Blog: `int (sec^2x)/(tan^2x+5tanx+6) dx` Use substitution and partial fractions to find the indefinite integral

Wednesday, 3 December 2014

$\displaystyle\int\frac{{{{\sec}}^{{2}}{x}}}{{{{\tan}}^{{2}}{x}+{5}{\tan{{x}}}+{6}}}{\left.{d}{x}\right.}$ Use substitution and partial fractions to find the indefinite integral

$\displaystyle\int\frac{{{{\sec}}^{{2}}{\left({x}\right)}}}{{{{\tan}}^{{2}}{\left({x}\right)}+{5}{\tan{{\left({x}\right)}}}+{6}}}{\left.{d}{x}\right.}$

Let's apply integral substitution: $\displaystyle{u}={\tan{{\left({x}\right)}}}$

$\displaystyle\Rightarrow{d}{u}={{\sec}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{1}}{{{{u}}^{{2}}+{5}{u}+{6}}}{d}{u}$

Now we have to write down integrand as sum of partial fraction function, but first we will have to factor the denominator,

$\displaystyle\frac{{1}}{{{{u}}^{{2}}+{5}{u}+{6}}}=\frac{{1}}{{{{u}}^{{2}}+{2}{u}+{3}{u}+{6}}}$

$\displaystyle=\frac{{1}}{{{u}{\left({u}+{2}\right)}+{3}{\left({u}+{2}\right)}}}$

$\displaystyle=\frac{{1}}{{{\left({u}+{2}\right)}{\left({u}+{3}\right)}}}$

Now let's create partial fraction template,

$\displaystyle\frac{{1}}{{{\left({u}+{2}\right)}{\left({u}+{3}\right)}}}=\frac{{A}}{{{u}+{2}}}+\frac{{B}}{{{u}+{3}}}$

Multiply the above equation by the denominator,

$\displaystyle\Rightarrow{1}={A}{\left({u}+{3}\right)}+{B}{\left({u}+{2}\right)}$

$\displaystyle{1}={A}{u}+{3}{A}+{B}{u}+{2}{B}$

$\displaystyle{1}={\left({A}+{B}\right)}{u}+{3}{A}+{2}{B}$

Equating the coefficients of the like terms,

$\displaystyle{A}+{B}={0}$ -----------------(1)

$\displaystyle{3}{A}+{2}{B}={1}$ -----------------(2)

From equation 1: $\displaystyle{A}=-{B}$

Substitute A in equation 2,

$\displaystyle{3}{\left(-{B}\right)}+{2}{B}={1}$

$\displaystyle-{3}{B}+{2}{B}={1}$

$\displaystyle\Rightarrow{B}=-{1}$

Plug...

$\displaystyle\int\frac{{{{\sec}}^{{2}}{\left({x}\right)}}}{{{{\tan}}^{{2}}{\left({x}\right)}+{5}{\tan{{\left({x}\right)}}}+{6}}}{\left.{d}{x}\right.}$

Let's apply integral substitution: $\displaystyle{u}={\tan{{\left({x}\right)}}}$

$\displaystyle\Rightarrow{d}{u}={{\sec}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{1}}{{{{u}}^{{2}}+{5}{u}+{6}}}{d}{u}$

Now we have to write down integrand as sum of partial fraction function, but first we will have to factor the denominator,

$\displaystyle\frac{{1}}{{{{u}}^{{2}}+{5}{u}+{6}}}=\frac{{1}}{{{{u}}^{{2}}+{2}{u}+{3}{u}+{6}}}$

$\displaystyle=\frac{{1}}{{{u}{\left({u}+{2}\right)}+{3}{\left({u}+{2}\right)}}}$

$\displaystyle=\frac{{1}}{{{\left({u}+{2}\right)}{\left({u}+{3}\right)}}}$

Now let's create partial fraction template,

$\displaystyle\frac{{1}}{{{\left({u}+{2}\right)}{\left({u}+{3}\right)}}}=\frac{{A}}{{{u}+{2}}}+\frac{{B}}{{{u}+{3}}}$

Multiply the above equation by the denominator,

$\displaystyle\Rightarrow{1}={A}{\left({u}+{3}\right)}+{B}{\left({u}+{2}\right)}$

$\displaystyle{1}={A}{u}+{3}{A}+{B}{u}+{2}{B}$

$\displaystyle{1}={\left({A}+{B}\right)}{u}+{3}{A}+{2}{B}$

Equating the coefficients of the like terms,

$\displaystyle{A}+{B}={0}$ -----------------(1)

$\displaystyle{3}{A}+{2}{B}={1}$ -----------------(2)

From equation 1: $\displaystyle{A}=-{B}$

Substitute A in equation 2,

$\displaystyle{3}{\left(-{B}\right)}+{2}{B}={1}$

$\displaystyle-{3}{B}+{2}{B}={1}$

$\displaystyle\Rightarrow{B}=-{1}$

Plug in the values in the partial fraction template,

$\displaystyle\frac{{1}}{{{\left({u}+{2}\right)}{\left({u}+{3}\right)}}}=\frac{{1}}{{{u}+{2}}}-\frac{{1}}{{{u}+{3}}}$

$\displaystyle\int\frac{{1}}{{{{u}}^{{2}}+{5}{u}+{6}}}{d}{u}=\int{\left(\frac{{1}}{{{u}+{2}}}-\frac{{1}}{{{u}+{3}}}\right)}{d}{u}$

Apply the sum rule,

$\displaystyle=\int\frac{{1}}{{{u}+{2}}}{d}{u}-\int\frac{{1}}{{{u}+{3}}}{d}{u}$

Use the common integral: $\displaystyle\int\frac{{1}}{{x}}{\left.{d}{x}\right.}={\ln}{\left|{x}\right|}$

$\displaystyle={\ln}{\left|{u}+{2}\right|}-{\ln}{\left|{u}+{3}\right|}$

Substitute back $\displaystyle{u}={\tan{{\left({x}\right)}}}$

and add a constant C to the solution,

$\displaystyle={\ln}{\left|{\tan{{\left({x}\right)}}}+{2}\right|}-{\ln}{\left|{\tan{{\left({x}\right)}}}+{3}\right|}+{C}$

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