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

Question

Evelyn Agard · Accepted Answer

#I=x/sqrt(a^2-x^2)-sin^-1(x/a)+c#

Here,

#I=intx^2/((a^2-x^2)^(3/2))dx#

Let ,#x=asint=>dx=acostdt#

#:.a^2-x^2=a^2-a^2sin^2t=a^2(1-sin^2t)=a^2cos^2t#

So,

#I=int(a^2sin^2t)/((a^2cos^2t)^(3/2))acostdt#

#=int(a^3sin^2tcost)/(a^3cos^3t)dt#

#=intsin^2t/cos^2tdt#

#=inttan^2tdt#

#=int(sec^2t-1)dt#

#=tant-t+c#

#=sint/cost-t+c#

#=sint/sqrt(1-sin^2t)-t+c#

Now, #x=asint=>sint=x/aand t=sin^-1(x/a)#

Hence,

#I=(x/a)/sqrt(1-(x^2/a^2))-sin^-1(x/a)+c#

#I=x/sqrt(a^2-x^2)-sin^-1(x/a)+c#

Ezra Adams · Answer

#x/(sqrt(a^2-x^2))-sin^(-1)(x/a)+c#

#intx^2/(a^2-x^2)^(3/2)dx#

rewrite as follows

#=int(x^2-a+a)/(a^2-x^2)^(3/2)dx#

#=color(red)(int(x^2-a^2)/(a^2-x^2)^(3/2)dx)+color(blue)(inta^2/(a^2-x^2)^(3/2)dx)---(1)#

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

from #(1)" " #the red integral

#color(red)(int(x^2-1)/(a^2-x^2)^(3/2)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

#color(red)(x=asinu=>dx=acosudu)#

#:. 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#

#color(blue)(x=asinu=>dx=acosudu)#

#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#

#"now "sinu=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

#x/(sqrt(a^2-x^2))-sin^(-1)(x/a)+c#

Aurora Alvarado · Answer

undefined To integrate $ \int \frac{x^2}{(a^2-x^2)^{3/2}} $ by trigonometric substitution, follow these steps:

1. Let $ x = a \sin \theta $.
2. Substitute $ dx = a \cos \theta \, d\theta $.
3. Rewrite $ (a^2 - x^2)^{3/2} $ as $ a^3 \cos^3 \theta $.
4. Substitute $ x $ and $ dx $ in terms of $ \theta $.
5. Simplify the integral and integrate with respect to $ \theta $.
6. Substitute back $ \sin \theta $ for $ x $ and simplify the result.

Let's proceed with these steps:

1. Let $ x = a \sin \theta $.
2. Substitute $ dx = a \cos \theta \, d\theta $.
3. Rewrite $ (a^2 - x^2)^{3/2} $ as $ a^3 \cos^3 \theta $.
4. Substitute $ x $ and $ dx $ in terms of $ \theta $:
$$ \int \frac{(a \sin \theta)^2}{(a^2 - (a \sin \theta)^2)^{3/2}} \cdot a \cos \theta \, d\theta $$
$$ = \int \frac{a^2 \sin^2 \theta}{(a^2 - a^2 \sin^2 \theta)^{3/2}} \cdot a \cos \theta \, d\theta $$
$$ = \int \frac{a^2 \sin^2 \theta}{(a^2 \cos^2 \theta)^{3/2}} \cdot a \cos \theta \, d\theta $$
$$ = \int \frac{a^2 \sin^2 \theta}{(a^3 \cos^3 \theta)} \cdot a \cos \theta \, d\theta $$

5. Simplify the integral and integrate with respect to $ \theta $:
$$ = \int \frac{a^3 \sin^2 \theta}{a^3 \cos^3 \theta} \, d\theta $$
$$ = \int \frac{\sin^2 \theta}{\cos^3 \theta} \, d\theta $$

6. Substitute back $ \sin \theta $ for $ x $ and simplify the result:
$$ = \int \frac{\sin^2 \theta}{\cos^3 \theta} \, d\theta $$
$$ = \int \frac{\tan^2 \theta}{\sec^3 \theta} \, d\theta $$

This integral can be solved using trigonometric identities or further simplification techniques.