How do you find the integral of #sin( x^(1/2) ) dx#?

Question

Scarlett Allison · Accepted Answer

#int sin(x^(1/2)) dx = 2sin(x^(1/2)) - 2x^(1/2)cos(x^(1/2)) + c#

Firstly, let #u = x^(1/2)#. By the power rule, #dx = 2x^(1/2) du = 2udu#.

By substituting #u# into the integral, we have:

#intsin(x^(1/2))dx = 2intusin(u)du#

We can solve this by integration by parts, which states that

#int fg' = fg - int f'g#

In our case, #f = u => f' = 1# and # g' = sin(u) => g = -cos(u)#.

#int usin(u)du = -ucos(u) - int-cos(u)du=#

#= -ucos(u) +intcos(u)du = -ucos(u)+sin(u) + C#

Therefore,

#2intusin(u)du = 2sin(u)-2ucos(u) + 2C#

A constant times another constant is still a constant, which we will call #c#.

#2C = c#

#2intusin(u)du = 2sin(u)-2ucos(u)+c#

By substituing #u=x^(1/2)# back, we have

#color(red)(intsin(x^(1/2))dx = 2sin(x^(1/2)) - 2x^(1/2)cos(x^(1/2)) +c)#.

Axel Alvarez · Answer

undefined To find the integral of sin( x^(1/2) ) dx, you can use a substitution. Let u = x^(1/2), then du = (1/2)x^(-1/2) dx = (1/(2x^(1/2))) dx. This means dx = 2x^(1/2) du. Substituting this into the integral, we get:

∫sin(u) * 2x^(1/2) du = 2∫x^(1/2)sin(u) du.

Now, we can integrate sin(u) to get -cos(u), so the integral becomes:

-2cos(u) + C = -2cos(x^(1/2)) + C.

Therefore, the integral of sin( x^(1/2) ) dx is -2cos( x^(1/2) ) + C, where C is the constant of integration.