What is the integral of 1/(sin^2(4x)) ?

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

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integral of 1/(sin^2(4x))

\int \frac{1}{sin ^{2} ( 4 x )} d x

- \frac{1}{4} cot (4 x) + C

Solution steps

\int \frac{1}{sin ^{2} ( 4 x )} d x

Rewrite using trig identities

= \int csc^{2} (4 x) d x

Apply u-substitution

: \int csc^{2} (u) \frac{1}{4} d u

= \int csc^{2} (u) \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int csc^{2} (u) d u

Use the common integral:

\int csc^{2} (u) d u = - cot (u)

= \frac{1}{4} (- cot (u))

Substitute back

u = 4 x

= \frac{1}{4} (- cot (4 x))

Simplify

= - \frac{1}{4} cot (4 x)

Add a constant to the solution

= - \frac{1}{4} cot (4 x) + C

x \to + 4 lim (\frac{∣ x - 4 ∣}{4 - x})

n = 0 \sum \infty 4^{n} x^{n}

\frac{\partial}{\partial y} (4 x + 3 y)

e^{x}, e^{5 x}, [0, 1]

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

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