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`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` .

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