Question

integral sin^2u du/sinx^3​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{\displaystyle\int\dfrac{sin^2u}{sinx^3}\;du}$

$\underline{\textbf{To evaluate:}}$

$\mathsf{\displaystyle\int\dfrac{sin^2u}{sinx^3}\;du}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{\displaystyle\int\,\dfrac{sin^2u}{sinx^3}\;du}$

$\mathsf{=\displaystyle\dfrac{1}{sinx^3}\int\;sin^2u\;du}$

$\mathsf{Using,}$

$\boxed{\mathsf{cos2A=1-2sin^2A\;\implies\;sin^2A=\dfrac{1-cos2A}{2}}}$

$\mathsf{=\displaystyle\dfrac{1}{sinx^3}\int\;\dfrac{1-cos2u}{2}\;du}$

$\mathsf{=\displaystyle\dfrac{1}{2\,sinx^3}\int\;(1-cos2u)\;du}$

$\mathsf{=\dfrac{1}{2\,sinx^3}\left[u-\dfrac{sin2u}{2}\right]+C}$

$\implies\boxed{\mathsf{\displaystyle\int\dfrac{sin^2u}{sinx^3}\;du=\dfrac{1}{2\,sinx^3}\left[u-\dfrac{sin2u}{2}\right]+C}}$

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