What is the integral of sec^6(x)tan^6(x) ?

The integral of sec^6(x)tan^6(x) is (tan^7(x))/7+(2tan^9(x))/9+(tan^{11}(x))/(11)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sec^6(x)tan^6(x)

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

\frac{tan ^{7} ( x )}{7} + \frac{2 tan ^{9} ( x )}{9} + \frac{tan ^{11} ( x )}{11} + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

: \int u^{6} (1 + u^{2})^{2} d u

= \int u^{6} (1 + u^{2})^{2} d u

Expand

u^{6} (1 + u^{2})^{2} : u^{6} + 2 u^{8} + u^{10}

= \int u^{6} + 2 u^{8} + u^{10} 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^{6} d u + \int 2 u^{8} d u + \int u^{10} d u

\int u^{6} d u = \frac{u ^{7}}{7}

\int 2 u^{8} d u = \frac{2 u ^{9}}{9}

\int u^{10} d u = \frac{u ^{11}}{11}

= \frac{u ^{7}}{7} + \frac{2 u ^{9}}{9} + \frac{u ^{11}}{11}

Substitute back

u = tan (x)

= \frac{tan ^{7} ( x )}{7} + \frac{2 tan ^{9} ( x )}{9} + \frac{tan ^{11} ( x )}{11}

Add a constant to the solution

= \frac{tan ^{7} ( x )}{7} + \frac{2 tan ^{9} ( x )}{9} + \frac{tan ^{11} ( x )}{11} + C

ln (x^{4} + 3)

\int_{0}^{1} 2 π (x) (e^{- x^{2}}) d x

(3 - x)^{2}

e^{x} - cot (x)

\frac{1}{( s ( s ^{2} + s + 1 ) )}

What is the integral of sec^6(x)tan^6(x) ?
The integral of sec^6(x)tan^6(x) is (tan^7(x))/7+(2tan^9(x))/9+(tan^{11}(x))/(11)+C