Question

$\int\frac{dx}{sin^2{(x)}cos^2{(x)}}$

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Answer 1

Answer:

$\sf I=tan\:x-cot\:x+C$

Step-by-step explanation:

Given:

$\sf \dfrac{1}{sin^2x\:cos^2x}$

To Find:

$\displaystyle \sf \int\limits {\dfrac{1}{sin^2x\:cos^2x} } \, dx$

Solution:

We know that,

sin²x + cos²x = 1

Hence,

$\displaystyle \sf \int\limits {\dfrac{1}{sin^2x\:cos^2x} } \, dx=\int\limits {\dfrac{sin^2x+cos^2x}{sin^2x\:cos^2x} } \, dx$

Now splitting the integral,

$\implies \displaystyle \sf \int\limits {\dfrac{sin^2x}{sin^2x\:cos^2x} } \, dx +\int\limits {\dfrac{cos^2x}{sin^2x\:cos^2x} } \, dx$

$\implies \displaystyle \sf \int\limits {\dfrac{1}{cos^2x} } \, dx +\int\limits {\dfrac{1}{sin^2x} } \, dx$

$\implies \displaystyle \sf \int\limits {sec^2x} \, dx +\int\limits {cosec^2x} \, dx$

We know that,

$\displaystyle \sf \int\limits {sec^2x} \, dx =tan\:x+C$

$\displaystyle \sf \int\limits {cosec^2x} \, dx=-cot\:x+C$

Substituting it we get,

$\sf \implies tan\:x-cot\:x+C$

This is the required integral.