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

Question

33342 views around the world

You can reuse this answer
Creative Commons License

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

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

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