int sin(sqrt(x)) dx Find the indefinite integral by using substitution followed by integration by parts.

To evaluate the given integral problem int sin(sqrt(x))dx using u-substitution, we may let:u = sqrt(x) .
 Square both sides of  u = sqrt(x) , we get: u^2 =x
Take the derivative of u^2 =x , we get: 2udu =dx .
Plug-in the values: u =sqrt(x) and dx = 2u du , we get:
 int sin(sqrt(x))dx =int sin(u)* 2u du
Apply the basic integration property: int c*f(x) dx = c int f(x) dx .
int sin(u)* 2u du =2int sin(u)* u du
Apply formula for integration by parts: int f*g'=f*g - int g*f' .
Let: f =u  then f' =du
       g' =sin(u) du then  g= -cos(u)
Note: From the table of integrals, we have int sin(theta) d theta= -cos(theta) +C .
Following the  formula for integration by parts, we set it up as:
2int sin(u)* u du= 2 * [ u *(-cos(u)) - int (-cos(u)) du]
                                 = 2 * [ -u cos(u)) + int (cos(u)) du]
                                 = 2 * [ -u cos(u)) + sin(u)]+C
                                 = -2ucos(u) +2sin(u) +C
Plug-in u=sqrt(x) on -2ucos(u) +2sin(u) +C , we get the complete indefinite integral as:
int sin(sqrt(x))dx=-2sqrt(x)cos(sqrt(x)) +2sin(sqrt(x)) +C .