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

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

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integral of tan^2(θ)sec^2(θ)

\int tan^{2} (θ) sec^{2} (θ) d θ

\frac{tan ^{3} ( θ )}{3} + C

Solution steps

\int tan^{2} (θ) sec^{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

\frac{\partial}{\partial x} (sin (2 x + y) sin (y))

\int tan^{3} (x) sec^{2 n + 1} (x) d x

\int x^{3} - 1 d x

\int (x + 3) (x^{2} + 6 x + 7)^{6} d x

f (x) = 2 x^{4} - 16 x^{2} + 6

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