What is the integral of tan^5(x)sec^2(x) ?

The integral of tan^5(x)sec^2(x) is (tan^6(x))/6+C

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integral of tan^5(x)sec^2(x)

\int tan^{5} (x) sec^{2} (x) d x

\frac{tan ^{6} ( x )}{6} + C

Solution steps

\int tan^{5} (x) sec^{2} (x) d x

Rewrite using trig identities

= \int (1 + tan^{2} (x)) tan^{5} (x) d x

Apply u-substitution

: \int u^{5} d u

= \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= \frac{u ^{6}}{6}

Substitute back

u = tan (x)

= \frac{tan ^{6} ( x )}{6}

Add a constant to the solution

= \frac{tan ^{6} ( x )}{6} + C

x \to - 3 lim (- - x + 4)

\frac{d y}{d x} = y ln (x)

\frac{d}{d x} (\frac{x}{x ^{2} + 6})

\int (1 - x) (e^{- x} + 1) d x

\frac{d}{d x} (ln (3 - x))

What is the integral of tan^5(x)sec^2(x) ?
The integral of tan^5(x)sec^2(x) is (tan^6(x))/6+C