# How do you integrate int x^2/(a^2-x^2)^(3/2) by trigonometric substitution?

Apr 5, 2018

$I = \frac{x}{\sqrt{{a}^{2} - {x}^{2}}} - {\sin}^{-} 1 \left(\frac{x}{a}\right) + c$

#### Explanation:

Here,

$I = \int {x}^{2} / \left({\left({a}^{2} - {x}^{2}\right)}^{\frac{3}{2}}\right) \mathrm{dx}$

Let ,$x = a \sin t \implies \mathrm{dx} = a \cos t \mathrm{dt}$

$\therefore {a}^{2} - {x}^{2} = {a}^{2} - {a}^{2} {\sin}^{2} t = {a}^{2} \left(1 - {\sin}^{2} t\right) = {a}^{2} {\cos}^{2} t$

So,

$I = \int \frac{{a}^{2} {\sin}^{2} t}{{\left({a}^{2} {\cos}^{2} t\right)}^{\frac{3}{2}}} a \cos t \mathrm{dt}$

$= \int \frac{{a}^{3} {\sin}^{2} t \cos t}{{a}^{3} {\cos}^{3} t} \mathrm{dt}$

$= \int {\sin}^{2} \frac{t}{\cos} ^ 2 t \mathrm{dt}$

$= \int {\tan}^{2} t \mathrm{dt}$

$= \int \left({\sec}^{2} t - 1\right) \mathrm{dt}$

$= \tan t - t + c$

$= \sin \frac{t}{\cos} t - t + c$

$= \sin \frac{t}{\sqrt{1 - {\sin}^{2} t}} - t + c$

Now, $x = a \sin t \implies \sin t = \frac{x}{a} \mathmr{and} t = {\sin}^{-} 1 \left(\frac{x}{a}\right)$

Hence,

$I = \frac{\frac{x}{a}}{\sqrt{1 - \left({x}^{2} / {a}^{2}\right)}} - {\sin}^{-} 1 \left(\frac{x}{a}\right) + c$

$I = \frac{x}{\sqrt{{a}^{2} - {x}^{2}}} - {\sin}^{-} 1 \left(\frac{x}{a}\right) + c$

Apr 5, 2018

$\frac{x}{\sqrt{{a}^{2} - {x}^{2}}} - {\sin}^{- 1} \left(\frac{x}{a}\right) + c$

#### Explanation:

$\int {x}^{2} / {\left({a}^{2} - {x}^{2}\right)}^{\frac{3}{2}} \mathrm{dx}$

rewrite as follows

$= \int \frac{{x}^{2} - a + a}{{a}^{2} - {x}^{2}} ^ \left(\frac{3}{2}\right) \mathrm{dx}$

$= \textcolor{red}{\int \frac{{x}^{2} - {a}^{2}}{{a}^{2} - {x}^{2}} ^ \left(\frac{3}{2}\right) \mathrm{dx}} + \textcolor{b l u e}{\int {a}^{2} / {\left({a}^{2} - {x}^{2}\right)}^{\frac{3}{2}} \mathrm{dx}} - - - \left(1\right)$

we will deal with each integral in turn, and leave the constant until the end

from $\left(1\right) \text{ }$the red integral

$\textcolor{red}{\int \frac{{x}^{2} - 1}{{a}^{2} - {x}^{2}} ^ \left(\frac{3}{2}\right) \mathrm{dx}}$

color(red)(I_1=-int(a^2-x^2)/(a^2-x^2)^(3/2)dx=-int1/(a^2-x^2)^(1/2)dx

we proceed by substitution

$\textcolor{red}{x = a \sin u \implies \mathrm{dx} = a \cos u \mathrm{du}}$

:. color(red)(I_1=-int1/(cancel((a^2(1-sin^2x))^(1/2)))xx cancel(acosu)du

color(red)(I_1=-intdu

color(red)(I_1=-u=-sin^(-1)(x/a)---(2)

now take the second (blue )integral from (2)" "~ and solve by substitution

color(blue)(I_2=inta^2/(a^2-x^2)^(3/2)dx

$\textcolor{b l u e}{x = a \sin u \implies \mathrm{dx} = a \cos u \mathrm{du}}$

color(blue)(I_2=inta^2/((a^2(1-sin^2u))^(3/2)) xx a cosudu

which simplifies to

color(blue)(intdu/cos^2u=intsec^2udu=tanu

$\text{now } \sin u = \frac{x}{a}$

:.tanu=(x/a)/(sqrt(a^2-x^2)/a)=x/(sqrt(a^2-x^2)

color(blue)(I_2=x/(sqrt(a^2-x^2))#

the final integral becomes

$\frac{x}{\sqrt{{a}^{2} - {x}^{2}}} - {\sin}^{- 1} \left(\frac{x}{a}\right) + c$