What is the integral of tan^8(θ)sec^2(θ) ?

The integral of tan^8(θ)sec^2(θ) is (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^8(θ)sec^2(θ)

\int tan^{8} (θ) sec^{2} (θ) d θ

\frac{tan ^{9} ( θ )}{9} + C

Solution steps

\int tan^{8} (θ) sec^{2} (θ) d θ

Rewrite using trig identities

= \int (1 + tan^{2} (θ)) tan^{8} (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 = tan (θ)

= \frac{tan ^{9} ( θ )}{9}

Add a constant to the solution

= \frac{tan ^{9} ( θ )}{9} + C

\int \frac{6}{e ^{4 x}} d x

\int \frac{2 x - 1}{x ^{2} - 2 x - 24} d x

\int (5 x^{3} + 3 cos (x)) d x

\int sin (17 x) d x

What is the integral of tan^8(θ)sec^2(θ) ?
The integral of tan^8(θ)sec^2(θ) is (tan^9(θ))/9+C