How do you find the antiderivative of $\int (sin (\sqrt{x})) d x$ $int (sin(sqrtx)) dx$ ?

Question

1558 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-26T21:55:41

$\int sin (\sqrt{x}) d x = - 2 \sqrt{x} cos (\sqrt{x}) + 2 sin (\sqrt{x}) + C$ $intsin(sqrtx)dx=-2sqrtxcos(sqrtx)+2sin(sqrtx)+C$

$I = \int sin (\sqrt{x}) d x$ $I=intsin(sqrtx)dx$

Let $t = \sqrt{x}$ $t=sqrtx$ . This implies that $x = t^{2}$ $x=t^2$ so $d x = 2 t d t$ $dx=2tdt$ .

Then:

$I = \int sin (t) (2 t d t) = \int 2 t sin (t) d t$ $I=intsin(t)(2tdt)=int2tsin(t)dt$

Now we will use integration by parts. Let:

${(u=2t,=>,du=2dt),(dv=sin(t)dt,=>,v=-cos(t)):}$

Then:

$I=uv-intvdu$

$I=-2tcos(t)-int(-cos(t))2dt$

$I=-2tcos(t)+2intcos(t)$

$I=-2tcos(t)+2sin(t)+C$

From $t=sqrtx$ :

$I=-2sqrtxcos(sqrtx)+2sin(sqrtx)+C$

How do you find the antiderivative of int (sin(sqrtx)) dx∫(sin(√x))dxint (sin(sqrtx)) dx?