Mathematics

$$\int {\cos e{c^m}x\cot x\,dx = } $$


ANSWER

$$\frac{{ - 1}}{{m\,{{\sin }^m}x}} + C$$


SOLUTION
$$\begin{array}{l}I = \int {{\rm{cose}}{{\rm{c}}^m}x\cot xdx} \\I = \int {{\rm{cose}}{{\rm{c}}^{m - 2}}{\rm{cose}}{{\rm{c}}^2}x\cot xdx} \\{\rm{Let}},\\\cot x = t\\ - {{\mathop{\rm cosc}\nolimits} ^2}xdx = dt\\1 + {\cot ^2}x = {{\mathop{\rm cosc}\nolimits} ^2}x\\{\left( {1 + {t^2}} \right)^{\frac{1}{2}}} = {\mathop{\rm cosc}\nolimits} x\\I = \int { - {{\left( {1 + {t^2}} \right)}^{\frac{{m - 2}}{2}}}dt} \\ = \frac{{ - 1}}{{m{{\sin }^m}x}} + C\end{array}$$
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Single Correct Medium Published on 17th 09, 2020
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