∫cosec2xdxcosec2x=sec2xtan2x ⟹ ∫sec2xtan2xdx,Let u = tanx∴du=sec2xdx ⟹ dx=dusec2xSubstituting the values above, we have∫1u2du=−u−1+c, Recall that u = tanx∴∫cosec2xdx=−cotx+c\displaystyle \int\cosec^2xdx\\ \cosec^2x= \frac{\sec^2x}{\tan^2x}\\ \implies \int \frac{\sec^2x}{\tan^2x}dx,\qquad \text{Let u = tanx}\\ \therefore du = \sec^2xdx \implies dx = \frac{du}{sec^2x}\\ \text{Substituting the values above, we have}\\ \int \frac{1}{u^2}du = -u^{-1}+c, \text{ Recall that u = tanx}\\ \therefore \int cosec^2xdx = -cotx + c∫cosec2xdxcosec2x=tan2xsec2x⟹∫tan2xsec2xdx,Let u = tanx∴du=sec2xdx⟹dx=sec2xduSubstituting the values above, we have∫u21du=−u−1+c, Recall that u = tanx∴∫cosec2xdx=−cotx+c
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