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

The integral of 1/(sin^2(x/2)cos^2(x/2)) is -4cot(x)+C

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

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

- 4 cot (x) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{8}{1 - cos ( 2 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 ( 2 x )} d x

Rewrite using trig identities

= 8 \cdot \int \frac{1}{2 sin ^{2} ( x )} d x

Take the constant out:

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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{sin ^{2} ( x )} d x

Use the common integral:

\int \frac{1}{sin ^{2} ( x )} d x = - cot (x)

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

Simplify

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

= - 4 cot (x)

Add a constant to the solution

= - 4 cot (x) + C

\frac{d}{d x} (ln (\frac{4 x}{x ^{2} + 5}))

\frac{d}{d x} (sin (x + y))

x \to 2 lim ((x^{2} - 4)^{- 1} sin (x - 2))

y^{^{''}} + 4 y = 6 sin (3 t)

\int 6 sin^{4} (x) cos (x) d x

What is the integral of 1/(sin^2(x/2)cos^2(x/2)) ?
The integral of 1/(sin^2(x/2)cos^2(x/2)) is -4cot(x)+C