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

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

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

- 2 cot (2 x) + C

Solution steps

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

Rewrite using trig identities

= \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

\int 12 x^{3} + sec (x) tan (x) d x

\int x^{3} (x^{2} + 1)^{2} d x

\int \frac{( x ^{3} + 3 x ^{6} )}{x ^{4}} d x

\int 7 x x^{2} + 1 d x

\int cos (5 t) + t^{4} d t

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