Mathematics

Evaluate: $$\displaystyle \int \dfrac{\sin x}{\sin 3x} dx$$


SOLUTION
$$ \int \frac{sinx}{sin3x}dx$$
$$ = \int \frac{sinx}{3sinx-4sin^{3}x}dx$$
$$ = \int \frac{dx}{3-4sin^{2}x}$$
$$ = \int \frac{dx}{\frac{3-4(1-cos2x)}{2}}$$
$$ = \int \frac{dx}{3-2(1-cos2x)}$$
$$ = \int \frac{dx}{3-2+2cox2x}$$
$$ = \int \frac{dx}{1+2cos2x}$$
$$ = \int \frac{dx}{1+\frac{2(1-t^{2})}{1+t^{2}}}$$
t = tanx
$$ dt = sec^{2}xdx \Rightarrow dx = \frac{dt}{1+t^{2}}$$
$$ = \int \frac{dt}{1+t^{2}+2-2t^{2}}$$
$$ = \int \frac{dt}{3-t^{2}}$$
$$ \int \frac{sinx}{sin3x}dx= \frac{1}{2\sqrt{3}}ln(\left | \frac{\sqrt{3}+tanx}{\sqrt{3}-tanx} \right |)+c$$

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Subjective Medium Published on 17th 09, 2020
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