What is the integral of tan^3(x)sec^8(x) ?

The integral of tan^3(x)sec^8(x) is (tan^4(x))/4+(tan^6(x))/2+(3tan^8(x))/8+(tan^{10}(x))/(10)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of tan^3(x)sec^8(x)

\int tan^{3} (x) sec^{8} (x) d x

\frac{tan ^{4} ( x )}{4} + \frac{tan ^{6} ( x )}{2} + \frac{3 tan ^{8} ( x )}{8} + \frac{tan ^{10} ( x )}{10} + C

Solution steps

\int tan^{3} (x) sec^{8} (x) d x

Rewrite using trig identities

= \int sec^{2} (x) (1 + tan^{2} (x))^{3} tan^{3} (x) d x

Apply u-substitution

= \int u^{3} (1 + u^{2})^{3} d u

Expand

u^{3} (1 + u^{2})^{3} : u^{3} + 3 u^{5} + 3 u^{7} + u^{9}

= \int u^{3} + 3 u^{5} + 3 u^{7} + u^{9} 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 u^{3} d u + \int 3 u^{5} d u + \int 3 u^{7} d u + \int u^{9} d u

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

\int 3 u^{5} d u = \frac{u ^{6}}{2}

\int 3 u^{7} d u = \frac{3 u ^{8}}{8}

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

= \frac{u ^{4}}{4} + \frac{u ^{6}}{2} + \frac{3 u ^{8}}{8} + \frac{u ^{10}}{10}

Substitute back

u = tan (x)

= \frac{tan ^{4} ( x )}{4} + \frac{tan ^{6} ( x )}{2} + \frac{3 tan ^{8} ( x )}{8} + \frac{tan ^{10} ( x )}{10}

Add a constant to the solution

= \frac{tan ^{4} ( x )}{4} + \frac{tan ^{6} ( x )}{2} + \frac{3 tan ^{8} ( x )}{8} + \frac{tan ^{10} ( x )}{10} + C

What is the integral of tan^3(x)sec^8(x) ?
The integral of tan^3(x)sec^8(x) is (tan^4(x))/4+(tan^6(x))/2+(3tan^8(x))/8+(tan^{10}(x))/(10)+C