Notes: `int cosx/sqrt(sin^2x+1) dx` Use integration tables to find the indefinite integral.

Tuesday, 12 November 2013

$\displaystyle\int\frac{{\cos{{x}}}}{\sqrt{{{{\sin}}^{{2}}{x}+{1}}}}{\left.{d}{x}\right.}$ Use integration tables to find the indefinite integral.

Recall that indefinite integral follows: $\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{C}$

where:

$\displaystyle{f{{\left({x}\right)}}}$ as the integrand function

$\displaystyle{F}{\left({x}\right)}$ as the antiderivative of $\displaystyle{f{{\left({x}\right)}}}$

$\displaystyle{C}$ as constant of integration.

To evaluate given integral problem: $\displaystyle\int\frac{{\cos{{\left({x}\right)}}}}{\sqrt{{{{\sin}}^{{2}}{\left({x}\right)}+{1}}}}{\left.{d}{x}\right.}$ or $\displaystyle\int\frac{{{\cos{{\left({x}\right)}}}{\left.{d}{x}\right.}}}{\sqrt{{{{\sin}}^{{2}}{\left({x}\right)}+{1}}}}$ , we may apply u-substitution by letting:

$\displaystyle{u}={\sin{{\left({x}\right)}}}$ then $\displaystyle{d}{u}={\cos{{\left({x}\right)}}}{\left.{d}{x}\right.}$ .

Plug-in the values , the integral becomes:

$\displaystyle\int\frac{{{\cos{{\left({x}\right)}}}{\left.{d}{x}\right.}}}{\sqrt{{{{\sin}}^{{2}}{\left({x}\right)}+{1}}}}=\int\frac{{{d}{u}}}{\sqrt{{{{u}}^{{2}}+{1}}}}$ or $\displaystyle\int\frac{{{d}{u}}}{\sqrt{{{{u}}^{{2}}+{{1}}^{{2}}}}}$

The integral resembles one of the formulas from the integration table for ...

Recall that indefinite integral follows: $\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{C}$

where:

$\displaystyle{f{{\left({x}\right)}}}$ as the integrand function

$\displaystyle{F}{\left({x}\right)}$ as the antiderivative of $\displaystyle{f{{\left({x}\right)}}}$

$\displaystyle{C}$ as constant of integration.

$\displaystyle{u}={\sin{{\left({x}\right)}}}$ then $\displaystyle{d}{u}={\cos{{\left({x}\right)}}}{\left.{d}{x}\right.}$ .

Plug-in the values , the integral becomes:

The integral resembles one of the formulas from the integration table for rational function with roots. We follow:

$\displaystyle\int\frac{{{\left.{d}{x}\right.}}}{\sqrt{{{{x}}^{{2}}+{{a}}^{{2}}}}}={\ln}{\left|{x}+\sqrt{{{{x}}^{{2}}+{{a}}^{{2}}}}\right|}+{C}$

By comparing $\displaystyle{{x}}^{{2}}+{{a}}^{{2}}$ with $\displaystyle{{u}}^{{2}}+{{1}}^{{2}}$ , we determine the corresponding values as: x=u and a=1.

Applying the values on the integral formula for rational function with roots, we get:

$\displaystyle\int\frac{{{d}{u}}}{\sqrt{{{{u}}^{{2}}+{{1}}^{{2}}}}}={\ln}{\left|{u}+\sqrt{{{{u}}^{{2}}+{{1}}^{{2}}}}\right|}+{C}$

$\displaystyle={\ln}{\left|{u}+\sqrt{{{{u}}^{{2}}+{1}}}\right|}+{C}$

Plug-in $\displaystyle{u}={\sin{{\left({x}\right)}}}$ on $\displaystyle{\ln}{\left|{u}+\sqrt{{{{u}}^{{2}}+{1}}}\right|}+{C}$ , we get the indefinite integral as:

$\displaystyle\int\frac{{\cos{{\left({x}\right)}}}}{\sqrt{{{{\sin}}^{{2}}{\left({x}\right)}+{1}}}}{\left.{d}{x}\right.}={\ln}{\left|{\sin{{\left({x}\right)}}}+\sqrt{{{{\sin}}^{{2}}{\left({x}\right)}+{1}}}\right|}+{C}$

Aside from this, we can also consider the another formula from integration table:

$\displaystyle\int\frac{{1}}{\sqrt{{{{u}}^{{2}}+{1}}}}{d}{u}={\arcsin{{h}}}{\left({u}\right)}+{C}$

Plug-in $\displaystyle{u}={\sin{{\left({x}\right)}}}$ on $\displaystyle{\arcsin{{h}}}{\left({u}\right)}+{C}$ , we get another form of indefinite integral as:

$\displaystyle\int\frac{{\cos{{\left({x}\right)}}}}{\sqrt{{{{\sin}}^{{2}}{\left({x}\right)}+{1}}}}{\left.{d}{x}\right.}={\arcsin{{h}}}{\left({\sin{{\left({x}\right)}}}\right)}+{C}$