Number 1 $\displaystyle \int y^2 \, {\rm csch} \, y^3 \, dy$

$\displaystyle = \frac{1}{3} \int {\rm csch} \, y^3 (3y^2\, dy)$

$\displaystyle = -\frac{1}{3} {\rm coth} \, y^3 + C$

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## Re: patulong po Hyperbolic and Inverse Hyperbolic functions...

Number 1$\displaystyle \int y^2 \, {\rm csch} \, y^3 \, dy$

$\displaystyle = \frac{1}{3} \int {\rm csch} \, y^3 (3y^2\, dy)$

$\displaystyle = -\frac{1}{3} {\rm coth} \, y^3 + C$