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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...

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    $\begingroup$ Try simpler: $t=\sqrt x$. $\endgroup$
    – Bernard
    Commented Feb 23, 2015 at 17:40

3 Answers 3

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$$\text{Put }u = \sqrt x \implies du = \frac 1{2\sqrt x}\,dx$$

$$2\int\frac{\sin\sqrt{x}}{2\sqrt{x}}dx \quad \text{ becomes } \quad2\int \sin u \,du= -2\cos u + c = -2\cos(\sqrt x) + c$$

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  • $\begingroup$ Thx, should have thought of it earlier) $\endgroup$ Commented Feb 23, 2015 at 18:11
  • $\begingroup$ You're welcome! $\endgroup$
    – amWhy
    Commented Feb 23, 2015 at 18:11
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Hint: Take $u = \sqrt{x}$ then $du = \frac{1}{2\sqrt{x}} dx \implies 2u \ du = dx$

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Hint: Put $t = \sqrt{x}$. Then, the integrand becomes: $2sin(t)$.

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