What is the integral of tan^5(pix)sec^4(pix) ?

The integral of tan^5(pix)sec^4(pix) is 1/pi ((tan^6(pix))/6+(tan^8(pix))/8)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of tan^5(pix)sec^4(pix)

\int tan^{5} (π x) sec^{4} (π x) d x

\frac{1}{π} (\frac{tan ^{6} ( π x )}{6} + \frac{tan ^{8} ( π x )}{8}) + C

Solution steps

\int tan^{5} (π x) sec^{4} (π x) d x

Apply u-substitution

= \int tan^{5} (u) sec^{4} (u) \frac{1}{π} d u

Take the constant out:

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

= \frac{1}{π} \cdot \int tan^{5} (u) sec^{4} (u) d u

Rewrite using trig identities

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

Apply u-substitution

= \frac{1}{π} \cdot \int v^{5} (1 + v^{2}) d v

Expand

v^{5} (1 + v^{2}) : v^{5} + v^{7}

= \frac{1}{π} \cdot \int v^{5} + v^{7} 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}{π} (\int v^{5} d v + \int v^{7} d v)

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

\int v^{7} d v = \frac{v ^{8}}{8}

= \frac{1}{π} (\frac{v ^{6}}{6} + \frac{v ^{8}}{8})

Substitute back

= \frac{1}{π} (\frac{tan ^{6} ( π x )}{6} + \frac{tan ^{8} ( π x )}{8})

Add a constant to the solution

= \frac{1}{π} (\frac{tan ^{6} ( π x )}{6} + \frac{tan ^{8} ( π x )}{8}) + C

What is the integral of tan^5(pix)sec^4(pix) ?
The integral of tan^5(pix)sec^4(pix) is 1/pi ((tan^6(pix))/6+(tan^8(pix))/8)+C