how to solve $\int\frac{\sin(x)-x\cos(x)}{x\sqrt{x\sin(x)}}dx$

Question

I was searching online for integral problem then I found this $$\int\frac{\sin(x)-x\cos(x)}{x\sqrt{x\sin(x)}}dx$$ which I couldn't solve my attempt is to use $y^2=x$ then $$I=2\int\frac{\sin(y^2)-y^2\cos(y^2)}{y^2\sqrt{\sin(y^2)}}dy$$ but here I have no idea what to do

Answer 1

Rewrite the integrand as

$$\begin{align*} & \int\frac{\sin(x)-x\cos(x)}{x\sqrt{x\sin(x)}} \, dx \\ &= \int \left(\frac{\sqrt{\sin x}}{x\sqrt x} - \frac{\cos x}{\sqrt{x \sin x}}\right) \, dx \\ &= \int \frac{\sin x-x\cos x}{x\sin x} \sqrt{\frac{\sin x}x} \, dx \end{align*}$$

Now substitute

$$u=\sqrt{\frac{\sin x}x} \implies du = -\frac12 \sqrt{\frac x{\sin x}} \frac{\sin x-x\cos x}{x^2}\,dx = -\frac12 \frac{\sin x-x\cos x}{x\sqrt{x\sin x}} \, dx$$

Answer 2

Noting that $$ \frac{d}{d x}\left(\frac{x}{\sin x}\right)=\frac{\sin x-x \cos x}{\sin ^2 x}, $$ we get $$ \begin{aligned} & I=\int \frac{\sin x-x \cos x}{\sin ^2 x} \cdot \frac{\sin ^2 x}{x \sqrt{x \sin x}} d x \\ & =\int\left(\frac{x}{\sin x}\right)^{-\frac{3}{2}} d\left(\frac{x}{\sin x}\right) \\ & =- 2\sqrt{\frac{\sin x}{x}} \\ & \end{aligned} $$

Answer 3

I randomly decided to see my old questions to see if I can answer them now

$$I:=\int \frac{\sin(x) -x \cos(x)}{x\sqrt{x \sin(x)}}dx$$ as I did before I will substitute $x=t^2$ then $dx =2tdt $ $$ I=\operatorname{sgn(x)} \int 2t \frac{\sin(t^2) -t^2 \cos(t^2)}{t^3\sqrt{\sin(t^2)}}dt= I=\operatorname{sgn(x)} \int 2 \frac{\sin(t^2) -t^2 \cos(t^2)}{t^2\sqrt{\sin(t^2)}}dt =2 I=\operatorname{sgn(x)} \left( \int \frac{\sqrt{\sin(t^2)}}{t^2} dt -\int \frac{ \cos(t^2) }{ \sqrt{\sin(t^2)} }dt \right)$$

$$ \text{since } \ \int \frac{\sqrt{\sin(t^2)}}{t^2} dt= - \frac{\sqrt{\sin(t^2)}}{t} + \int \frac{ \cos(t^2) }{ \sqrt{\sin(t^2)} }dt$$ $$\text{then} \ \ I= -2 \operatorname{sgn(x)} \frac{\sqrt{\sin(t^2)}}{t} +C=-2 \operatorname{sgn(x)} \frac{\sqrt{\sin(x)}}{\sqrt{x}} +C$$