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

The integral of sec^5(x)tan^7(x) is (sec^{11}(x)}{11}-\frac{sec^9(x))/3+(3sec^7(x))/7-(sec^5(x))/5+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{sec ^{11} ( x )}{11} - \frac{sec ^{9} ( x )}{3} + \frac{3 sec ^{7} ( x )}{7} - \frac{sec ^{5} ( x )}{5} + C

Solution steps

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

Rewrite using trig identities

= \int (- 1 + sec^{2} (x))^{3} tan (x) sec^{5} (x) d x

Apply u-substitution

: \int u^{4} (u^{2} - 1)^{3} d u

= \int u^{4} (u^{2} - 1)^{3} d u

Expand

u^{4} (u^{2} - 1)^{3} : u^{10} - 3 u^{8} + 3 u^{6} - u^{4}

= \int u^{10} - 3 u^{8} + 3 u^{6} - u^{4} 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 3 u^{8} d u + \int 3 u^{6} d u - \int u^{4} d u

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

\int 3 u^{8} d u = \frac{u ^{9}}{3}

\int 3 u^{6} d u = \frac{3 u ^{7}}{7}

\int u^{4} d u = \frac{u ^{5}}{5}

= \frac{u ^{11}}{11} - \frac{u ^{9}}{3} + \frac{3 u ^{7}}{7} - \frac{u ^{5}}{5}

Substitute back

u = sec (x)

= \frac{sec ^{11} ( x )}{11} - \frac{sec ^{9} ( x )}{3} + \frac{3 sec ^{7} ( x )}{7} - \frac{sec ^{5} ( x )}{5}

Add a constant to the solution

= \frac{sec ^{11} ( x )}{11} - \frac{sec ^{9} ( x )}{3} + \frac{3 sec ^{7} ( x )}{7} - \frac{sec ^{5} ( x )}{5} + C

\int \frac{x ^{2}}{( 4 + x ^{2} ) ^{\frac{3}{2}}} d x

\int \frac{4 x}{1 + x ^{3}} d x

\int \frac{3 x ^{2} + 2 x + 1}{( x + 1 ) ^{2} ( x ^{2} + 1 )} d x

\int - 4 x sin (x) d x

\int \frac{1}{45 - 20 x - 25 x ^{2}} d x

What is the integral of sec^5(x)tan^7(x) ?
The integral of sec^5(x)tan^7(x) is (sec^{11}(x)}{11}-\frac{sec^9(x))/3+(3sec^7(x))/7-(sec^5(x))/5+C