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Q.Evaluate: ∫sin⁡xsin⁡3x dx\displaystyle\int \dfrac{\sin x}{\sin 3x}\,dx

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2026Subjective· 3mImportance★★★★★
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Expand sin⁡3x\sin3x in terms of sin⁡x\sin x, cancel, then substitute t=tan⁡xt=\tan x.

Using sin⁡3x=3sin⁡x−4sin⁡3x=sin⁡x(3−4sin⁡2x)\sin3x=3\sin x-4\sin^3x=\sin x(3-4\sin^2x):

sin⁡xsin⁡3x=13−4sin⁡2x=13−4(1−cos⁡2x)=14cos⁡2x−1\frac{\sin x}{\sin3x}=\frac{1}{3-4\sin^2x}=\frac{1}{3-4(1-\cos^2x)}=\frac{1}{4\cos^2x-1}

Divide numerator and denominator by cos⁡2x\cos^2x:

14cos⁡2x−1=sec⁡2x4−sec⁡2x=sec⁡2x4−(1+tan⁡2x)=sec⁡2x3−tan⁡2x\frac{1}{4\cos^2x-1}=\frac{\sec^2x}{4-\sec^2x}=\frac{\sec^2x}{4-(1+\tan^2x)}=\frac{\sec^2x}{3-\tan^2x}

So ∫sin⁡xsin⁡3x dx=∫sec⁡2x3−tan⁡2x dx\displaystyle\int\frac{\sin x}{\sin3x}\,dx=\int\frac{\sec^2x}{3-\tan^2x}\,dx.

Let t=tan⁡xt=\tan x, dt=sec⁡2x dxdt=\sec^2x\,dx: …

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