# Find integral $\int\frac{\sin\sqrt{x}}{\sqrt{x}}dx$

Please help me to find indefinite integral: $$\int\frac{\sin\sqrt{x}}{\sqrt{x}}dx$$ Please suggest to me a way to do it.

I tried the substitution $t = \dfrac{1}{\sqrt{x}},\,$ so that $\,dt = -\dfrac{1}{2x^{3/2}}\,dx$.

But I don't know what to do next...

• Try simpler: $t=\sqrt x$. – Bernard Feb 23 '15 at 17:40

$$\text{Put }u = \sqrt x \implies du = \frac 1{2\sqrt x}\,dx$$
$$2\int\frac{\sin\sqrt{x}}{2\sqrt{x}}dx \quad \text{ becomes } \quad2\int \sin u \,du= -2\cos u + c = -2\cos(\sqrt x) + c$$
Hint: Take $u = \sqrt{x}$ then $du = \frac{1}{2\sqrt{x}} dx \implies 2u \ du = dx$
Hint: Put $t = \sqrt{x}$. Then, the integrand becomes: $2sin(t)$.