New Notes Blog: `int sinx/(cosx+cos^2x) dx` Use substitution and partial fractions to find the indefinite integral

Tuesday, 8 July 2014

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

$\displaystyle\int\frac{{\sin{{\left({x}\right)}}}}{{{\cos{{\left({x}\right)}}}+{{\cos}}^{{2}}{\left({x}\right)}}}{\left.{d}{x}\right.}$

Apply integral substitution: $\displaystyle{u}={\cos{{\left({x}\right)}}}$

$\displaystyle\Rightarrow{d}{u}=-{\sin{{\left({x}\right)}}}{\left.{d}{x}\right.}$

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

Take the constant out,

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

Now to compute the partial fraction expansion of a proper rational function, we have to factor out the denominator,

$\displaystyle=-{1}\int\frac{{1}}{{{u}{\left({u}+{1}\right)}}}{d}{u}$

Now let's create the partial fraction expansion,

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

Multiply the above equation by the denominator,

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

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

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

Equating the coefficients of the like terms,

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

$\displaystyle{A}={1}$

Plug in the value of A in equation 1,

$\displaystyle{1}+{B}={0}$

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

Plug in the values of...

$\displaystyle\int\frac{{\sin{{\left({x}\right)}}}}{{{\cos{{\left({x}\right)}}}+{{\cos}}^{{2}}{\left({x}\right)}}}{\left.{d}{x}\right.}$

Apply integral substitution: $\displaystyle{u}={\cos{{\left({x}\right)}}}$

$\displaystyle\Rightarrow{d}{u}=-{\sin{{\left({x}\right)}}}{\left.{d}{x}\right.}$

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

Take the constant out,

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

Now to compute the partial fraction expansion of a proper rational function, we have to factor out the denominator,

$\displaystyle=-{1}\int\frac{{1}}{{{u}{\left({u}+{1}\right)}}}{d}{u}$

Now let's create the partial fraction expansion,

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

Multiply the above equation by the denominator,

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

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

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

Equating the coefficients of the like terms,

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

$\displaystyle{A}={1}$

Plug in the value of A in equation 1,

$\displaystyle{1}+{B}={0}$

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

Plug in the values of A and B in the partial fraction expansion,

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

$\displaystyle=\frac{{1}}{{u}}-\frac{{1}}{{{u}+{1}}}$

$\displaystyle\int\frac{{1}}{{{u}{\left({u}+{1}\right)}}}{d}{u}=\int{\left(\frac{{1}}{{u}}-\frac{{1}}{{{u}+{1}}}\right)}{d}{u}$

Apply the sum rule,

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

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

$\displaystyle={\ln}{\left|{u}\right|}-{\ln}{\left|{u}+{1}\right|}$

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

$\displaystyle={\ln}{\left|{\cos{{\left({x}\right)}}}\right|}-{\ln}{\left|{\cos{{\left({x}\right)}}}+{1}\right|}$

$intsin(x)/(cos(x)+cos^2(x))dx=-1{ln|cos|x|-ln|cos(x)+1|}$

Simplify and add a constant C to the solution,

$\displaystyle={\ln}{\left|{\cos{{\left({x}\right)}}}+{1}\right|}-{\ln}{\left|{\cos{{\left({x}\right)}}}\right|}+{C}$

New Notes Blog

Tuesday, 8 July 2014

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

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Tuesday, 8 July 2014

Use substitution and partial fractions to find the indefinite integral

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Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

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

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?