What is the integral of sec^4(x)tan^5(x) ?

The integral of sec^4(x)tan^5(x) is (tan^6(x))/6+(tan^8(x))/8+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sec^4(x)tan^5(x)

\int sec^{4} (x) tan^{5} (x) d x

\frac{tan ^{6} ( x )}{6} + \frac{tan ^{8} ( x )}{8} + C

Solution steps

\int sec^{4} (x) tan^{5} (x) d x

Rewrite using trig identities

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

Apply u-substitution

: \int u^{5} (1 + u^{2}) d u

= \int u^{5} (1 + u^{2}) d u

Expand

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

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

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

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

= \frac{u ^{6}}{6} + \frac{u ^{8}}{8}

Substitute back

u = tan (x)

= \frac{tan ^{6} ( x )}{6} + \frac{tan ^{8} ( x )}{8}

Add a constant to the solution

= \frac{tan ^{6} ( x )}{6} + \frac{tan ^{8} ( x )}{8} + C

\int x y^{4} d x

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

\int \frac{x ^{4} + 8 x}{x ^{2}} d x

\int 8 x^{3} 2 x^{4} + 6 d x

What is the integral of sec^4(x)tan^5(x) ?
The integral of sec^4(x)tan^5(x) is (tan^6(x))/6+(tan^8(x))/8+C