What is the integral of sec^7(x)tan^5(x) ?

The integral of sec^7(x)tan^5(x) is (sec^{11}(x)}{11}-\frac{2sec^9(x))/9+(sec^7(x))/7+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sec^7(x)tan^5(x)

\int sec^{7} (x) tan^{5} (x) d x

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

Solution steps

\int sec^{7} (x) tan^{5} (x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (x))^{2} tan (x) sec^{7} (x) d x

Apply u-substitution

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

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

Expand

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

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

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

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

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

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

Substitute back

u = sec (x)

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

Add a constant to the solution

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

\int cos (x) + 18 x d x

\int \frac{1 + 2 x ^{2}}{x ^{4} ( 1 + x ^{2} )} d x

\int \frac{1}{s ^{\frac{1}{2}} - s ^{\frac{1}{4}}} d s

\int \frac{x ^{2} - 8 x - 4}{x ^{3} - 4 x} d x

What is the integral of sec^7(x)tan^5(x) ?
The integral of sec^7(x)tan^5(x) is (sec^{11}(x)}{11}-\frac{2sec^9(x))/9+(sec^7(x))/7+C