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3.2(A) · Q67

Q.Integrate: ∫sin⁡6xsin⁡10xsin⁡4x dx\int \dfrac{\sin 6x}{\sin 10x\sin 4x}\,dx

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Write sin⁡6x=sin⁡(10x−4x)=sin⁡10xcos⁡4x−cos⁡10xsin⁡4x\sin6x=\sin(10x-4x)=\sin10x\cos4x-\cos10x\sin4x.

Dividing by sin⁡10xsin⁡4x\sin10x\sin4x: sin⁡6xsin⁡10xsin⁡4x=cos⁡4xsin⁡4x−cos⁡10xsin⁡10x=cot⁡4x−cot⁡10x\dfrac{\sin6x}{\sin10x\sin4x}=\dfrac{\cos4x}{\sin4x}-\dfrac{\cos10x}{\sin10x}=\cot4x-\cot10x.

∫cot⁡4x dx=14ln⁡∣sin⁡4x∣\int\cot4x\,dx=\dfrac14\ln|\sin4x|, ∫cot⁡10x dx=110ln⁡∣sin⁡10x∣\int\cot10x\,dx=\dfrac{1}{10}\ln|\sin10x|. …

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