What is the integral of 18tan^4(x) ?

The integral of 18tan^4(x) is 18(1/3 tan^3(x)+x-tan(x))+C

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integral of 18tan^4(x)

\int 18 tan^{4} (x) d x

18 (\frac{1}{3} tan^{3} (x) + x - tan (x)) + C

Solution steps

\int 18 tan^{4} (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 18 \cdot \int tan^{4} (x) d x

Apply Integral Reduction

= 18 (\frac{tan ^{3} ( x )}{3} - \int tan^{2} (x) d x)

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

= 18 (\frac{tan ^{3} ( x )}{3} - (- x + tan (x)))

Simplify

18 (\frac{tan ^{3} ( x )}{3} - (- x + tan (x))) : 18 (\frac{1}{3} tan^{3} (x) + x - tan (x))

= 18 (\frac{1}{3} tan^{3} (x) + x - tan (x))

Add a constant to the solution

= 18 (\frac{1}{3} tan^{3} (x) + x - tan (x)) + C

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What is the integral of 18tan^4(x) ?
The integral of 18tan^4(x) is 18(1/3 tan^3(x)+x-tan(x))+C