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

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

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

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

\frac{tan ^{3} ( x )}{3} + C

Solution steps

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

Rewrite using trig identities

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

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 (x)

= \frac{tan ^{3} ( x )}{3}

Add a constant to the solution

= \frac{tan ^{3} ( x )}{3} + C

x = 0, y = 15 e^{- x^{2}}, [0, 1]

(3, - 6), (- 1, 2)

f (x) = 3 + cot (x) - 2 csc (x)

\frac{t - t}{t ^{\frac{1}{5}}}

y = x^{2 x + 1}

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