# How do you integrate int sqrt(4-sqrtx) using substitution?

Mar 28, 2018

$\int \sqrt{4 - \sqrt{x}} \mathrm{dx} = - \frac{16}{3} {\left(4 - \sqrt{x}\right)}^{\frac{3}{2}} + \frac{4}{5} {\left(4 - \sqrt{x}\right)}^{\frac{5}{2}} + C$

#### Explanation:

Let $u = 4 - \sqrt{x}$

Solving for $\sqrt{x}$ yields

$\sqrt{x} = 4 - u$

And,

$\mathrm{du} = - \frac{1}{2 \sqrt{x}} \mathrm{dx}$

Initially, it does not seem like $\mathrm{du}$ shows up in the integral at all. But with some manipulation and simplification, this substitution becomes valid:

$- 2 \sqrt{x} \mathrm{du} = \mathrm{dx}$

$- 2 \left(4 - u\right) \mathrm{du} = \mathrm{dx}$

Thus, our integral becomes

$\int \sqrt{u} \left(- 2\right) \left(4 - u\right) \mathrm{du}$

$- 2 \int {u}^{\frac{1}{2}} \left(4 - u\right) \mathrm{du}$

$- 2 \int \left(4 {u}^{\frac{1}{2}} - {u}^{\frac{3}{2}}\right) \mathrm{du} = - 2 \left(\left(4\right) \left(\frac{2}{3}\right) {u}^{\frac{3}{2}} - \frac{2}{5} {u}^{\frac{5}{2}}\right) + C = - 2 \left(\frac{8}{3} {u}^{\frac{3}{2}} - \frac{2}{5} {u}^{\frac{5}{2}}\right) + C = - \frac{16}{3} {u}^{\frac{3}{2}} + \frac{4}{5} {u}^{\frac{5}{2}} + C$

Rewriting in terms of $x$ yields

$\int \sqrt{4 - \sqrt{x}} \mathrm{dx} = - \frac{16}{3} {\left(4 - \sqrt{x}\right)}^{\frac{3}{2}} + \frac{4}{5} {\left(4 - \sqrt{x}\right)}^{\frac{5}{2}} + C$

Mar 28, 2018

Look to explanation

#### Explanation:

Notice how you can use a whole integrand substitution
$u = \setminus \sqrt{4 - \setminus \sqrt{x}} \setminus \to x = {\left(4 - {u}^{2}\right)}^{2}$
And we notice that
$\mathrm{dx} = - 4 u \left(4 - {u}^{2}\right) \mathrm{du}$
So therefore we see that
$\setminus \int \setminus \sqrt{4 - \setminus \sqrt{x}} \mathrm{dx} = \setminus \int u \setminus \cdot \left(- 4 u \left(4 - {u}^{2}\right)\right) \mathrm{du}$
This can be simplified to
$\setminus \int 4 {u}^{4} - 16 {u}^{2} \mathrm{du} = \frac{4}{5} {u}^{5} - \frac{16}{3} {u}^{3} + C$
And then you need to substitute the value of $u$ to obtain
$\setminus \int \setminus \sqrt{4 - \setminus \sqrt{x}} \mathrm{dx} = \frac{4}{5} {\left(\setminus \sqrt{4 - \setminus \sqrt{x}}\right)}^{5} - \frac{16}{3} {\left(\setminus \sqrt{4 - \setminus \sqrt{x}}\right)}^{3} + C$