How do you integrate $\int sin (\sqrt{x})$ $int sin(sqrtx)$ by integration by parts method?

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Answer 1 · 2016-09-02T17:19:26

$2 sin (\sqrt{x}) - 2 \sqrt{x} cos (\sqrt{x}) + C$ $2sin(sqrt(x))-2 sqrt(x) cos(sqrt(x))+C$

Making $x = y^{2}$ $x = y^2$ in $\int sin (\sqrt{x}) d x$ $int sin(sqrtx)dx$

after $d x = 2 y d y$ $dx = 2 y dy$ we have

$\int sin (\sqrt{x}) d x \equiv 2 \int y sin y d y$ $int sin(sqrtx)dx equiv 2inty sin y dy$

but $\frac{d}{d y} (y cos y) = cos y - y sin y$ $d/(dy)(y cos y) = cosy -y sin y$ so

$2 \int y sin y d y = 2 \int cos y d y - 2 y cos y = 2 sin y - 2 y cos y + C$ $2inty sin y dy=2int cos y dy -2y cos y = 2sin y -2y cos y + C$

Finally

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

How do you integrate int sin(sqrtx)∫sin(√x)int sin(sqrtx) by integration by parts method?