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3.2(A) · Q68

Q.Integrate: ∫sin⁡xcos⁡3x1+cos⁡2x dx\int \dfrac{\sin x\cos^3 x}{1+\cos 2x}\,dx

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Use 1+cos⁡2x=2cos⁡2x1+\cos2x=2\cos^2x.

sin⁡xcos⁡3x2cos⁡2x=sin⁡xcos⁡x2\dfrac{\sin x\cos^3x}{2\cos^2x}=\dfrac{\sin x\cos x}{2}.

∫sin⁡xcos⁡x2dx=14∫2sin⁡xcos⁡x dx=14∫sin⁡2x dx⋅(equivalently)=14sin⁡2x\int\dfrac{\sin x\cos x}{2}dx=\dfrac14\int2\sin x\cos x\,dx=\dfrac14\int\sin2x\,dx\cdot\text{(equivalently)}=\dfrac14\sin^2x …

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