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I want to understand integration by substition.

$$\int \frac{\sin\big({\sqrt{x}}\big)}{\sqrt{x}} dx$$

$u = \sqrt{x}$

$\frac{du}{dx} = \frac{1}{2\sqrt{x}}$

$du = \frac{1}{2 \sqrt{x}} dx$

$dx = 2 \sqrt{x} du$

If I insert this into the integral, we get

$$\int \frac{\sin \big(\sqrt{x} \big)}{\sqrt{x}}dx = \int \frac{\sin(u)}{u} dx = \int \frac{\sin(u)}{u} 2 \sqrt{x} du = 2 \int \sin(u) \sqrt{x} du\tag{1}$$

What am I misunderstanding?

Why do we get to $$2 \int \sin(u) du$$

instead of

$$\int \frac{\sin(u)}{u} dx$$

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    $\begingroup$ $\frac{1}{u}2\sqrt x = \frac{1}{{\sqrt x }}2\sqrt x = 2$, writing $d$ is meaningless. $\endgroup$
    – Gary
    Commented Jul 27, 2022 at 13:12
  • $\begingroup$ You can write: $$u =\sqrt x\implies dx=2\sqrt x du \implies dx=2udu$$ and then cancel the $u$ with the denominator $\endgroup$
    – Matteo
    Commented Jul 27, 2022 at 13:14

6 Answers 6

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Note in your last step in Eq.$(1)$

$$... \int \frac{\sin(u)}{u} 2 \sqrt{x} du= 2 \int \sin(u) \sqrt{x} du\tag{1}$$

Here you need to replace $\sqrt{x}$ by $u$, so you get

$$... = \int \frac{\sin(u)}{u} ~2u~ du$$

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You got to $2\int \sin(u)du$ as well:

$$\int \frac{\sin(\sqrt{x})}{\sqrt{x}}dx =(*)$$

$$u=\sqrt{x}$$ $$dx=2\sqrt{x}du=2udu$$

$$(*)=\int \frac{\sin(u)}{u}2udu=2\int \sin(u)du$$

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From $u=\sqrt{x}$ we get $x =u^2.$ Therefore $dx=2u\,du.$ Hence $$\int {\sin\sqrt{x}\over \sqrt{x}}\,dx=\int {\sin u\over u}\, 2u\,du=2\int {\sin u}\,du$$

While substituting it is convenient not to mix variables $x$ and $u$ in one indefinite integral. Keep them separated. Similarly for $x=\varphi(u)$ we have $dx=\varphi'(u)\,du$ and $$\int f(x)\,dx =\int f(\varphi(u))\,\varphi'(u)\,du$$

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The issue arrises when substituting in $du=2 \sqrt x dx$.

Since $u = \sqrt x$ you can say the denominator of $u$ and $2u$ will cancel to give $2$ and so will overall give you; the integral of $2\sin(u)$.

In your working, you remove the denominator of $u$ without canceling out the $\sqrt x$ which is what causes your problem.

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$\int \frac{\sin\big({\sqrt{x}}\big)}{\sqrt{x}} dx= -2\cos(x^\frac{1}{2})+c$ by the chain rule.

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$$ \begin{aligned} I & \stackrel{y=\sqrt x}{=} \int \frac{\sin y}{y}( 2 y d y)=2 \int \sin y d y=-2 \cos y+C =-2 \cos \sqrt{x}+C . \end{aligned} $$

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