How to integrate $\int x\sin {(\sqrt{x})}\, dx$ I tried using integration by parts twice, the same way we do for  $\int \sin {(\sqrt{x})}$
but in the second integral, I'm not getting an expression that is equal to  $\int x\sin {(\sqrt{x})}$.
I let $\sqrt x = t$ thus,
$$\int t^2 \cdot \sin({t})\cdot 2t dt =  2\int t^3\sin(t)dt = 2[(-\cos(t)\cdot t^3 + \int 3t^2\cos(t))] = 2[-\cos(t)\cdot t^3+(\sin(t)\cdot 3t^3 - \int 6t \cdot \sin(t))]]$$
which I can't find useful. 
 A: Yes, indeed, continue as you did in the comments, treating $\int 6t\sin t \,dt\;$ as a separate integral, use integration by parts, and add (or subtract, if appropriate) that result to your earlier work, and you will end with an expression with no integrals remaining!:
$$\int t^2 \cdot \sin({t})\cdot 2t dt = $$
$$= 2[-\cos(\sqrt x) \cdot x(\sqrt x) + \sin(\sqrt x)\cdot 3x -(\cos(\sqrt x)\cdot6\sqrt x+\sin(\sqrt x)\cdot \sqrt x + \cos (\sqrt x))] + C$$
after substituting $\sqrt x$ for $t$, though I'd suggest finding a way to simplify (combining like terms, etc.)
A: Just continue your path of partial integration with the last integral ? The last integral is purely a cosine which is integrable and yields your sollution.
A: you have done till this - $ 2[-\cos(t)\cdot t^3+(\sin(t)\cdot 3t^3 - \int 6t \cdot \sin(t))]]$
again use parts to get 
$ 2[-\cos(t)\cdot t^3+(\sin(t)\cdot 3t^3 -  6(t(-\cos t) +\sin(t))$ = 
$ 2[-\cos(t)\cdot t^3+(\sin(t)\cdot 3t^3 +  6(t(\cos t) -6\sin(t))$ .
giving you your answer .
just put $ \sqrt{x} = t $ and you are through.
A: Integrating by parts, we get 
if $n\ne-1,$
$$\int x^n\cos\sqrt xdx= \frac{x^{n+1}\cos\sqrt x}{n+1}+\frac1{2(n+1)}\int x^{n+\frac12}\sin\sqrt x dx$$
$$\int x^n\sin\sqrt xdx= \frac{x^{n+1}\sin\sqrt x}{n+1}-\frac1{2(n+1)}\int x^{n+\frac12}\cos\sqrt x dx$$
Putting $n=\frac12$ in the first integral, 
$$\int x^\frac12\cos\sqrt xdx= \frac{x^{\frac12+1}\cos\sqrt x}{\frac12+1}+\frac1{2(\frac12+1)}\int x\sin\sqrt x dx$$
$$\implies \int x\sin\sqrt x dx=3\int x^\frac12\cos\sqrt xdx- 2x^{\frac12+1}\cos\sqrt x$$
Putting $n=0$ in the second integral, 
$$\int \sin\sqrt xdx= \frac{x \sin\sqrt x}{1}-\frac1{2}\int x^{\frac12}\cos\sqrt x dx$$
$$\implies \int x^{\frac12}\cos\sqrt x dx= 2x \sin\sqrt x-2\int \sin\sqrt xdx$$
Now, $\int \sin\sqrt xdx$ can be found here
