What is the integral of (tan(x))^5(sec(x))^3 ?

The integral of (tan(x))^5(sec(x))^3 is (sec^7(x))/7-(2sec^5(x))/5+(sec^3(x))/3+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (tan(x))^5(sec(x))^3

\int (tan (x))^{5} (sec (x))^{3} d x

\frac{sec ^{7} ( x )}{7} - \frac{2 sec ^{5} ( x )}{5} + \frac{sec ^{3} ( x )}{3} + C

Solution steps

\int (tan (x))^{5} (sec (x))^{3} d x

Simplify

= \int tan^{5} (x) sec^{3} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= \int u^{6} - 2 u^{4} + u^{2} 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^{4} d u + \int u^{2} d u

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

\int 2 u^{4} d u = \frac{2 u ^{5}}{5}

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

= \frac{u ^{7}}{7} - \frac{2 u ^{5}}{5} + \frac{u ^{3}}{3}

Substitute back

u = sec (x)

= \frac{sec ^{7} ( x )}{7} - \frac{2 sec ^{5} ( x )}{5} + \frac{sec ^{3} ( x )}{3}

Add a constant to the solution

= \frac{sec ^{7} ( x )}{7} - \frac{2 sec ^{5} ( x )}{5} + \frac{sec ^{3} ( x )}{3} + C

\int x^{2} - 2 x - 24 d x

\int x - 2 x d x

\int cos^{16} (θ) sin (θ) d θ

\int \frac{2 x - 5}{( x - 1 ) ^{2} ( x + 2 )} d x

\int 6 x^{5} + 7 d x

What is the integral of (tan(x))^5(sec(x))^3 ?
The integral of (tan(x))^5(sec(x))^3 is (sec^7(x))/7-(2sec^5(x))/5+(sec^3(x))/3+C