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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- cot (4 x) + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{1}{8 ( 1 - cos ( u ) )} d u

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

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

Substitute back

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

Simplify

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

= - cot (4 x)

Add a constant to the solution

= - cot (4 x) + C

\int sec^{3} (x) - sec (x) d x

\int_{0}^{1} (\frac{9 t}{1 + t ^{2}}) d t

y^{^{'}} = x (3 - y)

\int (x^{3} (x^{2} - 2 x)^{3}) d x

y = \frac{1}{2} (e^{x} - e^{- x})

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