What is the integral of sec^9(θ)tan(θ) ?

The integral of sec^9(θ)tan(θ) is (sec^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 sec^9(θ)tan(θ)

\int sec^{9} (θ) tan (θ) d θ

\frac{sec ^{9} ( θ )}{9} + C

Solution steps

\int sec^{9} (θ) tan (θ) d θ

Apply u-substitution

: \int u^{8} d u

= \int u^{8} d u

Apply the Power Rule

: \frac{u ^{9}}{9}

= \frac{u ^{9}}{9}

Substitute back

u = sec (θ)

= \frac{sec ^{9} ( θ )}{9}

Add a constant to the solution

= \frac{sec ^{9} ( θ )}{9} + C

\int \frac{1}{x ^{2} - 2 x - 1} d x

\int 4 x + 16 d x

\int 4 x - x^{2} - 3 d x

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

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

What is the integral of sec^9(θ)tan(θ) ?
The integral of sec^9(θ)tan(θ) is (sec^9(θ))/9+C