What is the integral of 1/((sin(x)cos(x))^2) ?

The integral of 1/((sin(x)cos(x))^2) is -2cot(2x)+C

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integral of 1/((sin(x)cos(x))^2)

\int \frac{1}{( sin ( x ) cos ( x ) ) ^{2}} d x

- 2 cot (2 x) + C

Solution steps

\int \frac{1}{( sin ( x ) cos ( x ) ) ^{2}} d x

Simplify

\frac{1}{( sin ( x ) cos ( x ) ) ^{2}} : csc^{2} (x) sec^{2} (x)

= \int csc^{2} (x) sec^{2} (x) d x

Rewrite using trig identities

= \int \frac{8}{1 - cos ( 4 x )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int \frac{1}{1 - cos ( 4 x )} d x

Apply u-substitution

: \int \frac{1}{4 ( 1 - cos ( u ) )} d u

= 8 \cdot \int \frac{1}{4 ( 1 - cos ( u ) )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \frac{1}{4} \cdot \int \frac{1}{1 - cos ( u )} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= 8 \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 8 \cdot \frac{1}{4} (- \frac{1}{v})

Substitute back

= 8 \cdot \frac{1}{4} (- \frac{1}{tan ( \frac{4 x}{2} )})

Simplify

8 \cdot \frac{1}{4} (- \frac{1}{tan ( \frac{4 x}{2} )}) : - 2 cot (2 x)

= - 2 cot (2 x)

Add a constant to the solution

= - 2 cot (2 x) + C

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What is the integral of 1/((sin(x)cos(x))^2) ?
The integral of 1/((sin(x)cos(x))^2) is -2cot(2x)+C