What is the integral of tan^4(θ)sec^6(θ) ?

The integral of tan^4(θ)sec^6(θ) is (tan^5(θ))/5+(2tan^7(θ))/7+(tan^9(θ))/9+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of tan^4(θ)sec^6(θ)

\int tan^{4} (θ) sec^{6} (θ) d θ

\frac{tan ^{5} ( θ )}{5} + \frac{2 tan ^{7} ( θ )}{7} + \frac{tan ^{9} ( θ )}{9} + C

Solution steps

\int tan^{4} (θ) sec^{6} (θ) d θ

Rewrite using trig identities

= \int sec^{2} (θ) (1 + tan^{2} (θ))^{2} tan^{4} (θ) d θ

Apply u-substitution

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

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

Expand

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

= \int u^{4} + 2 u^{6} + u^{8} 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 2 u^{6} d u + \int u^{8} d u

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

\int 2 u^{6} d u = \frac{2 u ^{7}}{7}

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

= \frac{u ^{5}}{5} + \frac{2 u ^{7}}{7} + \frac{u ^{9}}{9}

Substitute back

u = tan (θ)

= \frac{tan ^{5} ( θ )}{5} + \frac{2 tan ^{7} ( θ )}{7} + \frac{tan ^{9} ( θ )}{9}

Add a constant to the solution

= \frac{tan ^{5} ( θ )}{5} + \frac{2 tan ^{7} ( θ )}{7} + \frac{tan ^{9} ( θ )}{9} + C

y = 5 x, y = 3 x^{2}

y^{^{'}} + 2 t y^{2} = 0

x \to 0 lim (- \frac{x}{1 + x ^{2}})

\int \frac{4 + x ^{2}}{x ^{6}} d x

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

What is the integral of tan^4(θ)sec^6(θ) ?
The integral of tan^4(θ)sec^6(θ) is (tan^5(θ))/5+(2tan^7(θ))/7+(tan^9(θ))/9+C