1.

Integrate this​

Answer»

{\displaystyle{\int \dfrac{\sec^2x}{\cosec^2x}dx}}

{\displaystyle{\int \dfrac{\sin^2x}{\cos^2x}dx}}

{\displaystyle{\int {\tan^2x•dx}}}

{\displaystyle{\boxed{\left(1+\tan^2x=\sec^2x\right)}}}

{\displaystyle{\int (\sec^2x-1)dx}}

{\displaystyle{\int \sec^2x \int 1•dx}}

{\displaystyle{\tan x-x+c}}


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