Question

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

Answer 1

`intsqrt(4-sqrtx)dx=-16/3(4-sqrtx)^(3/2)+4/5(4-sqrtx)^(5/2)+C`

Explanation:

Let `u=4-sqrt(x)`

Solving for `sqrtx` yields

`sqrtx=4-u`

And,

`du=-1/(2sqrtx)dx`

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

`-2sqrtxdu=dx`

`-2(4-u)du=dx`

Thus, our integral becomes

`intsqrt(u)(-2)(4-u)du`

`-2intu^(1/2)(4-u)du`

`-2int(4u^(1/2)-u^(3/2))du=-2((4)(2/3)u^(3/2)-2/5u^(5/2))+C=-2(8/3u^(3/2)-2/5u^(5/2))+C=-16/3u^(3/2)+4/5u^(5/2)+C`

Rewriting in terms of `x` yields

`intsqrt(4-sqrtx)dx=-16/3(4-sqrtx)^(3/2)+4/5(4-sqrtx)^(5/2)+C`

Answer 2

Look to explanation

Explanation:

Notice how you can use a whole integrand substitution
`u=\sqrt{4-\sqrt{x}}\tox=(4-u^2)^2`
And we notice that
`dx=-4u(4-u^2) du`
So therefore we see that
`\int\sqrt{4-\sqrt{x}}dx=\int u\cdot(-4u(4-u^2))du`
This can be simplified to
`\int4u^4-16u^2 du=4/5u^5-16/3u^3+C`
And then you need to substitute the value of `u` to obtain
`\int\sqrt{4-\sqrt{x}}dx=4/5(\sqrt{4-\sqrt{x}})^5-16/3(\sqrt{4-\sqrt{x}})^3+C`