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

The integral of sec(2x)tan^5(2x) is 1/2 sec(2x)-(sec^3(2x))/3+(sec^5(2x))/(10)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{2} sec (2 x) - \frac{sec ^{3} ( 2 x )}{3} + \frac{sec ^{5} ( 2 x )}{10} + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

\frac{( - 1 + u ^{2} ) ^{2}}{2} : \frac{1}{2} - u^{2} + \frac{u ^{4}}{2}

= \int \frac{1}{2} - u^{2} + \frac{u ^{4}}{2} 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{1}{2} d u - \int u^{2} d u + \int \frac{u ^{4}}{2} d u

\int \frac{1}{2} d u = \frac{1}{2} u

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

\int \frac{u ^{4}}{2} d u = \frac{u ^{5}}{10}

= \frac{1}{2} u - \frac{u ^{3}}{3} + \frac{u ^{5}}{10}

Substitute back

u = sec (2 x)

= \frac{1}{2} sec (2 x) - \frac{sec ^{3} ( 2 x )}{3} + \frac{sec ^{5} ( 2 x )}{10}

Add a constant to the solution

= \frac{1}{2} sec (2 x) - \frac{sec ^{3} ( 2 x )}{3} + \frac{sec ^{5} ( 2 x )}{10} + C

\int (e^{x y}) d y

(\frac{1}{tan ( x )})^{^{'}}

lo g_{5} (5 x^{3} + 2 x + 5)

y^{^{'}} = tan (t) y - 5 t

2916 x, 108 x^{2}

What is the integral of sec(2x)tan^5(2x) ?
The integral of sec(2x)tan^5(2x) is 1/2 sec(2x)-(sec^3(2x))/3+(sec^5(2x))/(10)+C