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

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

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integral of tan^3(x)sec^2(x)

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

\frac{tan ^{4} ( x )}{4} + C

Solution steps

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

Rewrite using trig identities

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

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 (x)

= \frac{tan ^{4} ( x )}{4}

Add a constant to the solution

= \frac{tan ^{4} ( x )}{4} + C

y = 15 - x^{2}, y = x^{2} - 3

\int_{1}^{5} \frac{1}{1 + 2 x} d x

\frac{\partial}{\partial x} (\frac{x ^{2} y}{x ^{4} + y ^{2}})

∣ x - 2 ∣, 0 = x - 3 y + 6

\int_{2}^{3} \frac{40}{3 - x} d x

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