Mathematics - Indefinite Integration Question with Solution | TestHub

Let $I=\int \frac{\sin x}{\sin 3x}dx$. We use the trigonometric identity $\sin 3x=3\sin x-4{\sin }^{3}x$.$I=\int \frac{\sin x}{\sin x(3-4{\sin }^{2}x)}dx=\int \frac{1}{3-4{\sin }^{2}x}dx$. Substitute ${\sin }^{2}x=1-{\cos }^{2}x$: $I=\int \frac{1}{3-4(1-{\cos }^{2}x)}dx=\int \frac{1}{3-4+4{\cos }^{2}x}dx=\int \frac{1}{4{\cos }^{2}x-1}dx$. Divide the numerator and denominator by ${\cos }^{2}x$: $I=\int \frac{{\sec }^{2}x}{4-{\sec }^{2}x}dx$. Replace ${\sec }^{2}x$ with $1+{\tan }^{2}x$: $I=\int \frac{{\sec }^{2}x}{4-(1+{\tan }^{2}x)}dx=\int \frac{{\sec }^{2}x}{3-{\tan }^{2}x}dx$. Let $t=\tan x$, so $dt={\sec }^{2}xdx$. The integral becomes $I=\int \frac{dt}{3-t^2}=\int \frac{dt}{(\sqrt{3}{)}^2-t^2}$. This is a standard integral form $\int \frac{1}{a^2-x^2}dx=\frac{1}{2a}\ln \left|\frac{a+x}{a-x}\right|+C$. Thus, $I=\frac{1}{2\sqrt{3}}\ln \left|\frac{\sqrt{3}+t}{\sqrt{3}-t}\right|+C$. Substituting back $t=\tan x$, we get $I=\frac{1}{2\sqrt{3}}\ln \left|\frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x}\right|+C$.