 fertilizeki

2022-01-01

Trigonometric integrals Evaluate the following integrals.
$\int {\mathrm{tan}}^{2}xdx$ zesponderyd

Step 1
Explanation:
Given that,
$\int {\mathrm{tan}}^{2}xdx$
We know that,
${\mathrm{tan}}^{2}x={\mathrm{sec}}^{2}x-1$
Now integrate
$\int {\mathrm{tan}}^{2}xdx=\int {\mathrm{sec}}^{2}x-1dx$
$\int {\mathrm{tan}}^{2}xdx=\int {\mathrm{sec}}^{2}xdx-\int 1dx$
$\int {\mathrm{tan}}^{2}xdx=\mathrm{tan}x-x+C$
Step 2
Hence,
The integral of $\int {\mathrm{tan}}^{2}xdx$ is $\mathrm{tan}x-x+C$ where C is integral constant. Jim Hunt

$\int {\mathrm{tan}}^{2}xdx$
Expand the expression
$\int {\mathrm{sec}\left(x\right)}^{2}-1dx$
$\int {\mathrm{sec}\left(x\right)}^{2}dx-\int 1dx$
Evaluate
$\mathrm{tan}\left(x\right)-x$
$\mathrm{tan}\left(x\right)-x+C$ Vasquez

$\int \mathrm{tan}\left(x{\right)}^{2}dx$
We make a trigonometric substitution: $\mathrm{tan}\left(x\right)=$ and then $dt=1/\left(1={t}^{2}\right)$
$\int \frac{{t}^{2}}{{t}^{2}+1}dt$
Simplify the expression: The
$\int \frac{{x}^{2}}{{x}^{2}+1}dx$
degree of the numerator P(x) is greater than or equal to the degree of the denominator Q(x), so we divide the polynomials.
$\frac{{x}^{2}}{{x}^{2}+1}=1+\frac{-1}{{x}^{2}+1}$
Integrating the integer part, we get:
$\int 1dx=x$
Integrating further, we get:
$\int \frac{-1}{{x}^{2}+1}dx=-\mathrm{arctan}\left(x\right)$
$x-\mathrm{arctan}\left(x\right)+C$
Returning to the change of variables $\left(t=\mathrm{tan}\left(x\right)\right)$, we get:
$I=\mathrm{tan}\left(x\right)-\mathrm{arctan}\left(\mathrm{tan}\left(x\right)\right)+C$

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