Simple Integration Question? (1 Viewer)

Drongoski · Jul 28, 2022

Someone asked me how to do:

$\int sin\sqrt{2x}dx$

Lith_30 · Jul 28, 2022

use substitution first of $s=\sqrt{x}$

$\\ds=\frac{dx}{2\sqrt{x}}\\\\2sds=dx$

then $\int{\sin(\sqrt{2x})dx}=2\int{s\sin(\sqrt{2}s)ds}$

then use integration by parts to get the answer

Drongoski · Jul 28, 2022

Earlier I did this in a similar way, using my crazy method:

$\int sin\sqrt{2x}dx = \frac {1}{2}\int sin\sqrt{2x}\cdot d(\sqrt{2x})^2 \\ \\ = \frac {1}{2}\int sin \sqrt{2x} \cdot 2\sqrt {2x} \cdot d\sqrt{2x}= -\int \sqrt{2x} \cdot dcos\sqrt{2x} \\ \\ =-\sqrt{2x}cos\sqrt{2x} + \int cos\sqrt{2x} \cdot d\sqrt {2x} $ (by Parts) $\\ \\ = -\sqrt{2x}cos \sqrt {2x} + sin \sqrt{2x} + C$

Simple Integration Question? (1 Viewer)

Drongoski

Well-Known Member

Lith_30

o_o

Drongoski

Well-Known Member

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