- #1

- 832

- 30

## Homework Statement

Well, the exercise asks me to solve the next integral using an adequate substitution.

[tex]\displaystyle\int_{}^{}\sqrt[ ]{4-x^2}dx[/tex]

## The Attempt at a Solution

[tex]\displaystyle\int_{}^{}\sqrt[ ]{4-x^2}dx[/tex]

What I did was:

[tex]x=2\sin\theta[/tex]

[tex]dx=2\cos\theta d\theta[/tex]

So, then I get

[tex]\displaystyle\int_{}^{}\sqrt[ ]{4-x^2}dx=4\displaystyle\int_{}^{}\cos^2\theta d\theta=cos\theta\sin\theta+\displaystyle\int_{}^{}\sin^2\theta d\theta=cos\theta\sin\theta+\displaystyle\int_{}^{}(1-\cos^2\theta) d\theta=[/tex]

[tex]=cos\theta\sin\theta+\displaystyle\int_{}^{}d\theta-\displaystyle\int_{}^{}\cos^2\theta[/tex]

[tex]\Rightarrow{5\displaystyle\int_{}^{}\cos^2\theta d\theta=cos\theta\sin\theta+\theta}\Rightarrow{\displaystyle\int_{}^{}\cos^2\theta d\theta=\displaystyle\frac{1}{5}cos\theta\sin\theta+\displaystyle\frac{\theta}{5}}[/tex]

[tex]\theta=\arcsin(\displaystyle\frac{x}{2})[/tex]

[tex]\sin\theta=(\displaystyle\frac{x}{2})[/tex]

[tex]\cos\theta=\sqrt[ ]{1-\sin^2\theta}[/tex]

[tex]\cos\theta=\sqrt[ ]{1-\displaystyle\frac{x^2}{4}}[/tex]

[tex]\displaystyle\int_{}^{}\sqrt[ ]{4-x^2}=\displaystyle\frac{x}{10}\sqrt[ ]{1-\displaystyle\frac{x^2}{4}}+\displaystyle\frac{\arcsin(\displaystyle\frac{x}{2})}{5}[/tex]

I think I've made a mistake, but I don't know in which step. Perhaps when I've get back to the original variables, or when I've solved by parts the integral of the square of the cosine.

Bye there, and thanks for posting.