Question

integration of sin6x/sin10x sin4x​

Answer 1

$\underline{\textsf{To find:}}$

$\mathsf{\displaystyle\int\,\dfrac{sin6x}{sin10x\,sin4x}\,dx}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\mathsf{\displaystyle\int\,\dfrac{sin6x}{sin10x\,sin4x}\,dx}$

$\mathsf{=\displaystyle\int\,\dfrac{sin(10x-4x)}{sin10x\,sin4x}\,dx}$

$\textsf{using the identity,}$

$\boxed{\mathsf{sin(A-B)=sinA\,cosB-cosA\,sinB}}$

$\mathsf{=\displaystyle\int\,\dfrac{sin10x\,cos4x-cos10x\,sin4x}{sin10x\,sin4x}\,dx}$

$\mathsf{=\displaystyle\int\,\dfrac{sin10x\,cos4x}{sin10x\,sin4x}\,dx-\int\,\dfrac{cos10x\,sin4x}{sin10x\,sin4x}\,dx}$

$\mathsf{=\displaystyle\int\,\dfrac{cos4x}{sin4x}\,dx-\int\,\dfrac{cos10x}{sin10x}\,dx}$

$\mathsf{=\displaystyle\int\,cot\,4x\,dx-\int\,cot\,10x\,dx}$

$\mathsf{=\dfrac{1}{4}\log|sin\,4x|-\dfrac{1}{10}\log|sin\,10x|+C}$

$\underline{\textsf{Answer:}}$

$\mathsf{\displaystyle\int\,\dfrac{sin6x}{sin10x\,sin4x}\,dx=\dfrac{1}{4}\log|sin\,4x|-\dfrac{1}{10}\log|sin\,10x|+C}$

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