What is the integral of tan^3(x)+tan^5(x) ?

The integral of tan^3(x)+tan^5(x) is 1/2 sec^2(x)-sec^2(x)+1/4 sec^4(x)+C

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integral of tan^3(x)+tan^5(x)

\int tan^{3} (x) + tan^{5} (x) d x

\frac{1}{2} sec^{2} (x) - sec^{2} (x) + \frac{1}{4} sec^{4} (x) + C

Solution steps

\int tan^{3} (x) + tan^{5} (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int tan^{3} (x) d x + \int tan^{5} (x) d x

\int tan^{3} (x) d x = - ln ∣ sec (x) ∣ + \frac{sec ^{2} ( x )}{2}

\int tan^{5} (x) d x = ln ∣ sec (x) ∣ - sec^{2} (x) + \frac{sec ^{4} ( x )}{4}

= - ln ∣ sec (x) ∣ + \frac{sec ^{2} ( x )}{2} + ln ∣ sec (x) ∣ - sec^{2} (x) + \frac{sec ^{4} ( x )}{4}

Simplify

- ln ∣ sec (x) ∣ + \frac{sec ^{2} ( x )}{2} + ln ∣ sec (x) ∣ - sec^{2} (x) + \frac{sec ^{4} ( x )}{4} : \frac{1}{2} sec^{2} (x) - sec^{2} (x) + \frac{1}{4} sec^{4} (x)

= \frac{1}{2} sec^{2} (x) - sec^{2} (x) + \frac{1}{4} sec^{4} (x)

Add a constant to the solution

= \frac{1}{2} sec^{2} (x) - sec^{2} (x) + \frac{1}{4} sec^{4} (x) + C

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What is the integral of tan^3(x)+tan^5(x) ?
The integral of tan^3(x)+tan^5(x) is 1/2 sec^2(x)-sec^2(x)+1/4 sec^4(x)+C