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

Question

Isabella Ackerman · Accepted Answer

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

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#

Ezra Adams · Answer

Look to explanation

Notice how you can use a whole integrand substitution #u=\sqrt{4-\sqrt{x}} ox=(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#

Evelyn Agard · Answer

undefined To integrate $ \int \sqrt{4 - \sqrt{x}} $ using substitution, let $ u = \sqrt{x} $. Then, $ du = \frac{1}{2\sqrt{x}}dx $. After some manipulation, the integral becomes $ \int \frac{2u^2}{\sqrt{4 - u}} du $. This can be further simplified using another substitution, $ v = 4 - u $, which results in $ dv = -du $. Finally, the integral transforms into $ -\int \frac{2(4-v)}{\sqrt{v}} dv $. This integral can be computed using basic integration techniques, resulting in $ -8\sqrt{v} + 4v\sqrt{v} + C $, where $ C $ is the constant of integration. Substituting back the expressions for $ u $ and $ v $, we get the final answer: $ -8\sqrt{\sqrt{x}} + 4(4 - \sqrt{x})\sqrt{\sqrt{x}} + C $.