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

The integral of tan^5(2x)sec^2(2x) is (sec^6(2x)}{12}-\frac{sec^4(2x))/4+(sec^2(2x))/4+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{sec ^{6} ( 2 x )}{12} - \frac{sec ^{4} ( 2 x )}{4} + \frac{sec ^{2} ( 2 x )}{4} + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

Expand

\frac{u ( u ^{2} - 1 ) ^{2}}{2} : \frac{u ^{5}}{2} - u^{3} + \frac{u}{2}

= \int \frac{u ^{5}}{2} - u^{3} + \frac{u}{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{u ^{5}}{2} d u - \int u^{3} d u + \int \frac{u}{2} d u

\int \frac{u ^{5}}{2} d u = \frac{u ^{6}}{12}

\int u^{3} d u = \frac{u ^{4}}{4}

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

= \frac{u ^{6}}{12} - \frac{u ^{4}}{4} + \frac{u ^{2}}{4}

Substitute back

u = sec (2 x)

= \frac{sec ^{6} ( 2 x )}{12} - \frac{sec ^{4} ( 2 x )}{4} + \frac{sec ^{2} ( 2 x )}{4}

Add a constant to the solution

= \frac{sec ^{6} ( 2 x )}{12} - \frac{sec ^{4} ( 2 x )}{4} + \frac{sec ^{2} ( 2 x )}{4} + C

What is the integral of tan^5(2x)sec^2(2x) ?
The integral of tan^5(2x)sec^2(2x) is (sec^6(2x)}{12}-\frac{sec^4(2x))/4+(sec^2(2x))/4+C