Question

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

Answer 1

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

Explanation:

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`

Answer 2

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

Explanation:

`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`