What is the integral of tan^3(6x)sec^{10}(6x) ?

The integral of tan^3(6x)sec^{10}(6x) is (sec^{12}(6x))/(72)-(sec^{10}(6x))/(60)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of tan^3(6x)sec^{10}(6x)

\int tan^{3} (6 x) sec^{10} (6 x) d x

\frac{sec ^{12} ( 6 x )}{72} - \frac{sec ^{10} ( 6 x )}{60} + C

Solution steps

\int tan^{3} (6 x) sec^{10} (6 x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (6 x)) tan (6 x) sec^{10} (6 x) d x

Apply u-substitution

: \int \frac{u ^{9} ( u ^{2} - 1 )}{6} d u

= \int \frac{u ^{9} ( u ^{2} - 1 )}{6} d u

Expand

\frac{u ^{9} ( u ^{2} - 1 )}{6} : \frac{u ^{11}}{6} - \frac{u ^{9}}{6}

= \int \frac{u ^{11}}{6} - \frac{u ^{9}}{6} d u

Apply the Sum Rule:

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

= \int \frac{u ^{11}}{6} d u - \int \frac{u ^{9}}{6} d u

\int \frac{u ^{11}}{6} d u = \frac{u ^{12}}{72}

\int \frac{u ^{9}}{6} d u = \frac{u ^{10}}{60}

= \frac{u ^{12}}{72} - \frac{u ^{10}}{60}

Substitute back

u = sec (6 x)

= \frac{sec ^{12} ( 6 x )}{72} - \frac{sec ^{10} ( 6 x )}{60}

Add a constant to the solution

= \frac{sec ^{12} ( 6 x )}{72} - \frac{sec ^{10} ( 6 x )}{60} + C

\frac{d ^{2}}{d x ^{2}} (x (2 x + 3)^{5})

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

h \to 0 lim (3 - 4 h)

\frac{d}{d x} (- 600 x^{- 2})

x, x, 0, 2

What is the integral of tan^3(6x)sec^{10}(6x) ?
The integral of tan^3(6x)sec^{10}(6x) is (sec^{12}(6x))/(72)-(sec^{10}(6x))/(60)+C