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

The integral of sec^2(θ)tan^2(θ) is (tan^3(θ))/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^2(θ)tan^2(θ)

\int sec^{2} (θ) tan^{2} (θ) d θ

\frac{tan ^{3} ( θ )}{3} + C

Solution steps

\int sec^{2} (θ) tan^{2} (θ) d θ

Rewrite using trig identities

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

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

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

Substitute back

u = tan (θ)

= \frac{tan ^{3} ( θ )}{3}

Add a constant to the solution

= \frac{tan ^{3} ( θ )}{3} + C

y^{^{''}} + 62 y^{^{'}} + 18 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\int_{- 1}^{1} 4 - x^{2} d x

\frac{d y}{d x} = x (e + y)

y = - x^{2}

\int cosh (x) d x

What is the integral of sec^2(θ)tan^2(θ) ?
The integral of sec^2(θ)tan^2(θ) is (tan^3(θ))/3+C