Question

Integral by substitution trigonometric ? a- ) int x/sqrt(9-x²)

I can not completely resolve this issue, please someone give me a hint how to solve!

Answer 1

`intx/sqrt(9-x^2)dx=-sqrt(9-x^2)+C`

Explanation:

So, we want

`intx/sqrt(9-x^2)dx`

In general, if we encounter an integral with the root `sqrt(a^2-x^2),` we can make the trigonometric substitution

`x=asintheta`

(We could use cosine, but using sine avoids unnecessary negative signs popping up.)

Here, `a^2=9, a=3, x=3sintheta, dx=3costhetad theta`

Rewrite with the substitution:

`int(9sinthetacosthetad theta)/sqrt(9-9sin^2theta)`

Simplify:

`int(9sinthetacosthetad theta)/sqrt(9(1-sin^2theta))=3int(sinthetacosthetad theta)/sqrt(1-sin^2theta)`

Recall the identity

`sin^2theta+cos^2theta=1`

This tells us that `1-sin^2theta=cos^2theta.` Apply this identity to our integral:

`3int(sinthetacosthetad theta)/sqrt(cos^2theta)`

`sqrt(cos^2theta)=|costheta|=costheta` -- We drop the absolute value bars and assume we're positive.

`3int(sinthetacancelcostheta)/(cancelcostheta)d theta`

`3intsinthetad theta=-3costheta+C`

We need to go back to `x.` Now, recall `x=3sintheta`. Therefore, `sintheta=x/3`. Let's once again use the identity `sin^2theta+cos^2theta=1`:

`x^2/9+cos^2theta=9/9`

`cos^2theta=(9-x^2)/9`

`costheta=sqrt(9-x^2)/3`

So, our integral becomes

`intx/sqrt(9-x^2)dx=-sqrt(9-x^2)+C`

Answer 2

`intx/sqrt(9-x^2)dx=-sqrt(9-x^2)+C`

Explanation:

The question is to evaluate:

`intx/sqrt(9-x^2)dx`

Draw a right-angle triangle with an angle `theta`.

Label the side opposite `theta` as `x`.
Label the hypotenuse as `3`.

Now examine the triangle and notice that `sintheta=x/3`, and so:

`x=3sintheta`
`dx=3costheta` `d` `theta`

Also notice that the side of the triangle adjacent to `theta` must be `sqrt(9-x^2)` by the Pythagorean Theorem. Therefore:

`costheta=sqrt(9-x^2)/3`

`sqrt(9-x^2)=3costheta`

Now rewrite the integral in terms of `theta`:

`int` `((3sintheta))/((3costheta))` `(3costheta` `d` `theta)`

Now simplify:

`int` `((3sintheta))/(cancel((3costheta)))` `cancel((3costheta` `d` `theta)`

`int` `3sintheta` `d` `theta =-3costheta+C`

Now put the expression back in terms of `x`:

`=-sqrt(9-x^2)+C`