Question

integral of cot square x dx​

Answer 1

Answer:

x+cotx+C. As we know that sin^2x+cos^2x=1

Answer 2

The required answer is  $-\cot x-x+c$.

Step-by-step explanation:

Consider the provided integral $\int \cot^2x dx$

Use the identity: $1+\cot^2x=\csc^2x$

$\int \cot^2x dx=\int (\csc^2x-1) dx$

$\int \cot^2x dx=\int \csc^2x \, dx-\int1 dx$

Use the formula; $\int \csc^2x= -\cot(x)$

$\int \cot^2x dx=-\cot x-x+c$

Hence, the required answer is  $-\cot x-x+c$.

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