What is the integral of 9tan^5(x) ?

The integral of 9tan^5(x) is 9(ln|sec(x)|-sec^2(x)+(sec^4(x))/4)+C

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integral of 9tan^5(x)

\int 9 tan^{5} (x) d x

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

Solution steps

\int 9 tan^{5} (x) d x

Take the constant out:

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

= 9 \cdot \int tan^{5} (x) d x

Rewrite using trig identities

= 9 \cdot \int (- 1 + sec^{2} (x))^{2} tan (x) d x

Apply u-substitution

: \int \frac{( - 1 + u ^{2} ) ^{2}}{u} d u

= 9 \cdot \int \frac{( - 1 + u ^{2} ) ^{2}}{u} d u

Expand

\frac{( - 1 + u ^{2} ) ^{2}}{u} : \frac{1}{u} - 2 u + u^{3}

= 9 \cdot \int \frac{1}{u} - 2 u + u^{3} d u

Apply the Sum Rule:

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

= 9 (\int \frac{1}{u} d u - \int 2 u d u + \int u^{3} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int 2 u d u = u^{2}

\int u^{3} d u = \frac{u ^{4}}{4}

= 9 (ln ∣ u ∣ - u^{2} + \frac{u ^{4}}{4})

Substitute back

u = sec (x)

= 9 (ln ∣ sec (x) ∣ - sec^{2} (x) + \frac{sec ^{4} ( x )}{4})

Add a constant to the solution

= 9 (ln ∣ sec (x) ∣ - sec^{2} (x) + \frac{sec ^{4} ( x )}{4}) + C

e^{- (\frac{1}{x + 1})}

x \to 0 lim (- (\frac{1}{x} + 1)^{x})

y \to 0 lim (sin (3 y))

x \to 1 - lim (2 + x)

\frac{d}{d x} (\frac{\frac{d}{d x} ( 15 e ^{4 x} )}{8 x ^{2}})

What is the integral of 9tan^5(x) ?
The integral of 9tan^5(x) is 9(ln|sec(x)|-sec^2(x)+(sec^4(x))/4)+C