Mathematics

Evaluate :
$$\int \dfrac{1}{\sin^2x \cos^2x}dx$$


ANSWER

$$-2\cot 2x +C$$


SOLUTION
$$I=\int \dfrac{1}{\sin^2x \cos^2x}dx$$

$$I=\int \dfrac{4}{(2\sin x \cos x)^2}dx$$

$$I=4 \int \dfrac{1}{\sin^22x}dx$$

$$I=4\int \csc^2 2x \ dx=-\dfrac{4}{2} \cot 2x +C=-2\cot 2x +C$$
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Single Correct Medium Published on 17th 09, 2020
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