What is the integral of sec^2(θ)tan^3(θ) ?

The integral of sec^2(θ)tan^3(θ) is (tan^4(θ))/4+C

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integral of sec^2(θ)tan^3(θ)

\int sec^{2} (θ) tan^{3} (θ) d θ

\frac{tan ^{4} ( θ )}{4} + C

Solution steps

\int sec^{2} (θ) tan^{3} (θ) d θ

Rewrite using trig identities

= \int (1 + tan^{2} (θ)) tan^{3} (θ) d θ

Apply u-substitution

: \int u^{3} d u

= \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= \frac{u ^{4}}{4}

Substitute back

u = tan (θ)

= \frac{tan ^{4} ( θ )}{4}

Add a constant to the solution

= \frac{tan ^{4} ( θ )}{4} + C

\int \frac{( x + 2 )}{x} d x

x \to 0 lim (x^{2} ln ∣ x ∣)

\frac{\partial}{\partial x} (y^{x^{y}})

\frac{d}{d x} (e^{3 x} cos (4 x))

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{- 4} e^{2 t}

What is the integral of sec^2(θ)tan^3(θ) ?
The integral of sec^2(θ)tan^3(θ) is (tan^4(θ))/4+C