What is the integral of (cos(39x))/(sin^2(39x)) ?

The integral of (cos(39x))/(sin^2(39x)) is -1/39 csc(39x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (cos(39x))/(sin^2(39x))

\int \frac{cos ( 39 x )}{sin ^{2} ( 39 x )} d x

- \frac{1}{39} csc (39 x) + C

Solution steps

\int \frac{cos ( 39 x )}{sin ^{2} ( 39 x )} d x

Use the basic trigonometric identity:

\frac{cos ( x )}{sin ( x )} = cot (x)

= \int \frac{cot ( 39 x )}{sin ( 39 x )} d x

Apply u-substitution

: \int \frac{cot ( u )}{39 sin ( u )} d u

= \int \frac{cot ( u )}{39 sin ( u )} d u

Take the constant out:

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

= \frac{1}{39} \cdot \int \frac{cot ( u )}{sin ( u )} d u

Rewrite using trig identities

= \frac{1}{39} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= \frac{1}{39} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{39} (- \frac{1}{v})

Substitute back

= \frac{1}{39} (- \frac{1}{sin ( 39 x )})

Simplify

\frac{1}{39} (- \frac{1}{sin ( 39 x )}) : - \frac{1}{39} csc (39 x)

= - \frac{1}{39} csc (39 x)

Add a constant to the solution

= - \frac{1}{39} csc (39 x) + C

\int \frac{3 x ^{2} - 5 x - 4}{x ^{3} - x} d x

\int x^{13} d x

\int 5 x^{2} (2 x^{3} + 6)^{8} d x

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

What is the integral of (cos(39x))/(sin^2(39x)) ?
The integral of (cos(39x))/(sin^2(39x)) is -1/39 csc(39x)+C