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

The integral of sec^3(x)tan^3(x) is (sec^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 sec^3(x)tan^3(x)

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

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

= \frac{u ^{5}}{5} - \frac{u ^{3}}{3}

Substitute back

u = sec (x)

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

Add a constant to the solution

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

y^{^{'}} + (1 - \frac{1}{x}) y = x^{2}

\frac{3}{x - 2}

\int \frac{x ^{2}}{( x ^{2} + 9 ) ^{3}} d x

x \to 4 lim (x)

x (\frac{d y}{d x}) + 4 y = x^{3} - x

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