What is the integral of tan^5(6x)sec^2(6x) ?

The integral of tan^5(6x)sec^2(6x) is 1/36 tan^6(6x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of tan^5(6x)sec^2(6x)

\int tan^{5} (6 x) sec^{2} (6 x) d x

\frac{1}{36} tan^{6} (6 x) + C

Solution steps

\int tan^{5} (6 x) sec^{2} (6 x) d x

Apply u-substitution

: \int tan^{5} (u) sec^{2} (u) \frac{1}{6} d u

= \int tan^{5} (u) sec^{2} (u) \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int tan^{5} (u) sec^{2} (u) d u

Rewrite using trig identities

= \frac{1}{6} \cdot \int (1 + tan^{2} (u)) tan^{5} (u) d u

Apply u-substitution

: \int v^{5} d v

= \frac{1}{6} \cdot \int v^{5} d v

Apply the Power Rule

: \frac{v ^{6}}{6}

= \frac{1}{6} \cdot \frac{v ^{6}}{6}

Substitute back

= \frac{1}{6} \cdot \frac{tan ^{6} ( 6 x )}{6}

Simplify

\frac{1}{6} \cdot \frac{tan ^{6} ( 6 x )}{6} : \frac{1}{36} tan^{6} (6 x)

= \frac{1}{36} tan^{6} (6 x)

Add a constant to the solution

= \frac{1}{36} tan^{6} (6 x) + C

\frac{d}{d x} (\frac{1}{9 x ^{\frac{8}{9}}})

f (x) = ln (5) x^{4}

\frac{d}{d x} (\frac{2}{( x + 2 ) ^{3}})

\frac{\partial}{\partial x} (7 - x^{2} + x y - 2 y^{2})

\frac{d}{d x} (6 x^{- 2})

What is the integral of tan^5(6x)sec^2(6x) ?
The integral of tan^5(6x)sec^2(6x) is 1/36 tan^6(6x)+C