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