# Question regarding integration with substition for $\int \frac{\sin({\sqrt{x}})}{\sqrt{x}}dx$

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 $$2 \int \sin(u) du$$

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

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

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

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

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

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

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

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

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 $$\int f(x)\,dx =\int f(\varphi(u))\,\varphi'(u)\,du$$

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.

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

\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}