1.7 Integrals resulting in inverse trigonometric functions  (Page 2/3)

${\displaystyle {\int }_0^{\sqrt{3}\text{/}2}\frac{dx}{\sqrt{1-x^2}}}$

${\displaystyle {\int }_{\mathrm{-1}\text{/}2}^{1\text{/}2}\frac{dx}{\sqrt{1-x^2}}}$

${\displaystyle {\int }_{\sqrt{3}}^1\frac{dx}{\sqrt{1+x^2}}}$

${\displaystyle {\int }_{1\text{/}\sqrt{3}}^{\sqrt{3}}\frac{dx}{1+x^2}}$

${\displaystyle {\int }_1^{\sqrt{2}}\frac{dx}{\left|x\right|\sqrt{x^2-1}}}$

${\displaystyle {\int }_1^{2\text{/}\sqrt{3}}\frac{dx}{\left|x\right|\sqrt{x^2-1}}}$

${\displaystyle \int \frac{dx}{\sqrt{9-x^2}}}$

${\displaystyle \int \frac{dx}{\sqrt{1-16x^2}}}$

${\displaystyle \int \frac{dx}{9+x^2}}$

${\displaystyle \int \frac{dx}{25+16x^2}}$

${\displaystyle \int \frac{dx}{\left|x\right|\sqrt{x^2-9}}}$

${\displaystyle \int \frac{dx}{\left|x\right|\sqrt{4x^2-16}}}$

Explain the relationship $\text{−}{\text{cos}}^{\mathrm{-1}}t+C={\displaystyle \int \frac{dt}{\sqrt{1-t^2}}={\text{sin}}^{\mathrm{-1}}t+C.}$ Is it true, in general, that ${\text{cos}}^{\mathrm{-1}}t=\text{−}{\text{sin}}^{\mathrm{-1}}t?$

Explain the relationship ${\text{sec}}^{\mathrm{-1}}t+C={\displaystyle \int \frac{dt}{\left|t\right|\sqrt{t^2-1}}=\text{−}{\text{csc}}^{\mathrm{-1}}t+C.}$ Is it true, in general, that ${\text{sec}}^{\mathrm{-1}}t=\text{−}{\text{csc}}^{\mathrm{-1}}t?$

Explain what is wrong with the following integral: ${\displaystyle {\int }_1^2\frac{dt}{\sqrt{1-t^2}}.}$

Explain what is wrong with the following integral: ${\displaystyle {\int }_{\mathrm{-1}}^1\frac{dt}{\left|t\right|\sqrt{t^2-1}}.}$

[T] ${\displaystyle \int \frac{1}{\sqrt{9-x^2}}dx}$ over $\left[\mathrm{-3},3\right]$

[T] ${\displaystyle \int \frac{9}{9+x^2}dx}$ over $\left[\mathrm{-6},6\right]$

[T] ${\displaystyle \int \frac{\text{cos}x}{4+{\text{sin}}^{2}x}dx}$ over $\left[\mathrm{-6},6\right]$

[T] ${\displaystyle \int \frac{e^x}{1+e^{2x}}dx}$ over $\left[\mathrm{-6},6\right]$

${\displaystyle \int \frac{{\text{sin}}^{\mathrm{-1}}tdt}{\sqrt{1-t^2}}}$

${\displaystyle \int \frac{dt}{{\text{sin}}^{\mathrm{-1}}t\sqrt{1-t^2}}}$

${\displaystyle \int \frac{{\text{tan}}^{\mathrm{-1}}\left(2t\right)}{1+4t^2}dt}$

${\displaystyle \int \frac{t{\text{tan}}^{\mathrm{-1}}\left(t^2\right)}{1+t^4}dt}$

${\displaystyle \int \frac{{\text{sec}}^{\mathrm{-1}}\left(\frac{t}{2}\right)}{\left|t\right|\sqrt{t^2-4}}dt}$

${\displaystyle \int \frac{t{\text{sec}}^{\mathrm{-1}}\left(t^2\right)}{t^2\sqrt{t^4-1}}dt}$

[T] ${\displaystyle \int \frac{1}{x\sqrt{x^2-4}}dx}$ over $\left[2,6\right]$

[T] ${\displaystyle \int \frac{1}{\left(2x+2\right)\sqrt{x}}dx}$ over $\left[0,6\right]$

[T] ${\displaystyle \int \frac{\left(\text{sin}x+x\text{cos}x\right)}{1+x^2{\text{sin}}^{2}x}dx}$ over $\left[\mathrm{-6},6\right]$