How to integrate $\int x\sin {(\sqrt{x})}\, dx$

Question

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.

Answer 1

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.)

Answer 2

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.

Answer 3

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.

Answer 4

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