What is the integral of tan^4(x)+tan^6(x) ?

The integral of tan^4(x)+tan^6(x) is (tan^5(x))/5+C

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integral of tan^4(x)+tan^6(x)

\int tan^{4} (x) + tan^{6} (x) d x

\frac{tan ^{5} ( x )}{5} + C

Solution steps

\int tan^{4} (x) + tan^{6} (x) d x

Simplify

tan^{4} (x) + tan^{6} (x) : tan^{4} (x) sec^{2} (x)

= \int tan^{4} (x) sec^{2} (x) d x

Rewrite using trig identities

= \int (1 + tan^{2} (x)) tan^{4} (x) d x

Apply u-substitution

: \int u^{4} d u

= \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= \frac{u ^{5}}{5}

Substitute back

u = tan (x)

= \frac{tan ^{5} ( x )}{5}

Add a constant to the solution

= \frac{tan ^{5} ( x )}{5} + C

h \to 0 lim (\frac{1}{5 - h})

\frac{e ^{- 2 s}}{s ( s - 1 )}

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

\int 13 tan (x) sec^{4} (x) d x

(x^{7} + 5 x)^{\frac{1}{3}}

What is the integral of tan^4(x)+tan^6(x) ?
The integral of tan^4(x)+tan^6(x) is (tan^5(x))/5+C