The Notes: `int sin(omega)/(cos^3(omega)) d omega` Evaluate the integral

Friday, December 27, 2013

$\displaystyle\int\frac{{\sin{{\left(\omega\right)}}}}{{{{\cos}}^{{3}}{\left(\omega\right)}}}{d}\omega$ Evaluate the integral

$\displaystyle\int\frac{{{\sin{{\left(\omega\right)}}}}}{{{{\cos}}^{{3}}{\left(\omega\right)}}}{d}\omega$

To solve, apply u-substitution method. So, let u be:

$\displaystyle{u}={\cos{{\left(\omega\right)}}}$

The, differentiate u.

$\displaystyle{d}{u}=-{\sin{{\left(\omega\right)}}}{d}\omega$

Since the sine function present in the integrand is positive, divide both sides by -1.

$\displaystyle\frac{{{d}{u}}}{{-{1}}}=\frac{{-{\sin{{\left(\omega\right)}}}{d}\omega}}{{-{1}}}$

$\displaystyle-{d}{u}={\sin{{\left(\omega\right)}}}{d}\omega$

Plugging them, the integral becomes:

$\displaystyle=\int\frac{{-{d}{u}}}{{{u}}^{{3}}}$

$\displaystyle=\int-{{u}}^{{-{3}}}{d}{u}$

To take the integral of this, apply the formula $\displaystyle\int\frac{{{u}}^{{{n}+{1}}}}{{{n}+{1}}}+{C}$ .

$\displaystyle=-\frac{{{u}}^{{-{2}}}}{{-{2}}}+{C}$

$\displaystyle=\frac{{{u}}^{{-{2}}}}{{2}}+{C}$

$\displaystyle=\frac{{1}}{{{2}{{u}}^{{2}}}}+{C}$

And, substitute back $\displaystyle{u}={\cos{{\left(\omega\right)}}}$ .

$\displaystyle=\frac{{1}}{{{2}{{\cos}}^{{2}}{\left(\omega\right)}}}+{C}$

Therefore, $\displaystyle\int\frac{{{\sin{{\left(\omega\right)}}}}}{{{{\cos}}^{{3}}{\left(\omega\right)}}}{d}\omega=\frac{{1}}{{{2}{{\cos}}^{{2}}{\left(\omega\right)}}}+{C}$ .

The Notes

Friday, December 27, 2013

$\displaystyle\int\frac{{\sin{{\left(\omega\right)}}}}{{{{\cos}}^{{3}}{\left(\omega\right)}}}{d}\omega$ Evaluate the integral

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