What is the integral of 1/(sin^4(3x)cos^2(3x)) ?

The integral of 1/(sin^4(3x)cos^2(3x)) is -1/3 (-tan(3x)+2cot(3x)+1/3 cot^3(3x))+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^4(3x)cos^2(3x))

\int \frac{1}{sin ^{4} ( 3 x ) cos ^{2} ( 3 x )} d x

- \frac{1}{3} (- tan (3 x) + 2 cot (3 x) + \frac{1}{3} cot^{3} (3 x)) + C

Solution steps

\int \frac{1}{sin ^{4} ( 3 x ) cos ^{2} ( 3 x )} d x

Rewrite using trig identities

= \int \frac{sec ^{2} ( 3 x )}{sin ^{4} ( 3 x )} d x

Use the basic trigonometric identity:

\frac{1}{sin ( x )} = csc (x)

= \int csc^{4} (3 x) sec^{2} (3 x) d x

Apply u-substitution

: \int csc^{4} (u) sec^{2} (u) \frac{1}{3} d u

= \int csc^{4} (u) sec^{2} (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int csc^{4} (u) sec^{2} (u) d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int \frac{1}{sin ^{4} ( u ) cos ^{2} ( u )} d u

Apply Trigonometric Substitution

: \int - \frac{( 1 + v ^{2} ) ^{2}}{v ^{2}} d v

= \frac{1}{3} \cdot \int - \frac{( 1 + v ^{2} ) ^{2}}{v ^{2}} d v

Take the constant out:

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

= \frac{1}{3} (- \int \frac{( 1 + v ^{2} ) ^{2}}{v ^{2}} d v)

Expand

\frac{( 1 + v ^{2} ) ^{2}}{v ^{2}} : \frac{1}{v ^{2}} + 2 + v^{2}

= \frac{1}{3} (- \int \frac{1}{v ^{2}} + 2 + v^{2} d v)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (- (\int \frac{1}{v ^{2}} d v + \int 2 d v + \int v^{2} d v))

\int \frac{1}{v ^{2}} d v = - \frac{1}{v}

\int 2 d v = 2 v

\int v^{2} d v = \frac{v ^{3}}{3}

= \frac{1}{3} (- (- \frac{1}{v} + 2 v + \frac{v ^{3}}{3}))

Substitute back

= \frac{1}{3} (- (- \frac{1}{cot ( 3 x )} + 2 cot (3 x) + \frac{cot ^{3} ( 3 x )}{3}))

Simplify

\frac{1}{3} (- (- \frac{1}{cot ( 3 x )} + 2 cot (3 x) + \frac{cot ^{3} ( 3 x )}{3})) : - \frac{1}{3} (- tan (3 x) + 2 cot (3 x) + \frac{1}{3} cot^{3} (3 x))

= - \frac{1}{3} (- tan (3 x) + 2 cot (3 x) + \frac{1}{3} cot^{3} (3 x))

Add a constant to the solution

= - \frac{1}{3} (- tan (3 x) + 2 cot (3 x) + \frac{1}{3} cot^{3} (3 x)) + C

\frac{x ^{2}}{x ^{2} - 1}

x \to 0 lim (\frac{6}{x ^{2}})

cos (θ) - θ sin (θ)

\frac{d}{d x} (\frac{a x + b}{c x - a})

x^{2} - 3, y = 1

What is the integral of 1/(sin^4(3x)cos^2(3x)) ?
The integral of 1/(sin^4(3x)cos^2(3x)) is -1/3 (-tan(3x)+2cot(3x)+1/3 cot^3(3x))+C