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3.2(B) · Q88

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

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Since sin⁡3x=3sin⁡x−4sin⁡3x=sin⁡x(3−4sin⁡2x)\sin3x=3\sin x-4\sin^3x=\sin x(3-4\sin^2x), we get sin⁡xsin⁡3x=13−4sin⁡2x\dfrac{\sin x}{\sin3x}=\dfrac{1}{3-4\sin^2x}.

Divide top and bottom by cos⁡2x\cos^2x: 13−4sin⁡2x=sec⁡2x3sec⁡2x−4tan⁡2x=sec⁡2x3(1+tan⁡2x)−4tan⁡2x=sec⁡2x3−tan⁡2x\dfrac{1}{3-4\sin^2x}=\dfrac{\sec^2x}{3\sec^2x-4\tan^2x}=\dfrac{\sec^2x}{3(1+\tan^2x)-4\tan^2x}=\dfrac{\sec^2x}{3-\tan^2x}. …

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