Kye

2021-10-12

Evaluate the following integrals.
$\int {\mathrm{sec}}^{-2}x{\mathrm{tan}}^{3}xdx$

cheekabooy

The given integral is
$I=\int {\mathrm{sec}}^{-2}x{\mathrm{tan}}^{3}xdx$
We have:
$I=\int {\mathrm{sec}}^{-2}x{\mathrm{tan}}^{3}xdx$
$I=\int \frac{{\mathrm{tan}}^{3}x}{{\mathrm{sec}}^{2}x}dx$
$I=\int \frac{{\mathrm{sin}}^{3}x{\mathrm{cos}}^{2}x}{{\mathrm{cos}}^{3}x}dx$
$I=\int \frac{{\mathrm{sin}}^{3}x}{\mathrm{cos}x}dx$
$I=\int \frac{\mathrm{sin}x{\mathrm{sin}}^{2}x}{\mathrm{cos}x}dx$
let $\mathrm{cos}x=t{\mathrm{cos}}^{2}x={t}^{2}⇒1-{\mathrm{sin}}^{2}x={t}^{2};{\mathrm{sin}}^{2}x=1-{t}^{2}$
$-\mathrm{sin}xdx=dt$
$⇒I=-\int \frac{\left(1-{t}^{2}\right)}{t}dt$
$⇒I=-\int \left(\frac{1}{t}-\frac{{t}^{2}}{t}\right)dt$
$⇒I=\int \left(t-\frac{1}{t}\right)dt$
$⇒I=\int tdt-\int \frac{1}{t}dt$
( using $\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C$ and $\int \frac{1}{x}dx=\mathrm{ln}|x|+C$)
$⇒I=\frac{{t}^{2}}{2}-\mathrm{ln}|\left(t\right)|+C$
$⇒I=\frac{{\mathrm{cos}}^{2}x}{2}-\mathrm{ln}|\mathrm{cos}x|+C$

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