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

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Answer 1 · 2018-04-05T22:03:00

#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)#.