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

The integral of (tan(x))^3 is 1/2 sec^2(x)+ln|cos(x)|+C

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integral of (tan(x))^3

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

\frac{1}{2} sec^{2} (x) + ln ∣ cos (x) ∣ + C

Solution steps

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

Rewrite using trig identities

= \int \frac{( 1 - cos ^{2} ( x ) ) sin ( x )}{cos ^{3} ( x )} d x

Apply u-substitution

: \int - \frac{1 - u ^{2}}{u ^{3}} d u

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

Expand

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

= \int - \frac{1}{u ^{3}} + \frac{1}{u} 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

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

= - (- \frac{1}{2 u ^{2}}) + ln ∣ u ∣

Substitute back

u = cos (x)

= - (- \frac{1}{2 cos ^{2} ( x )}) + ln ∣ cos (x) ∣

Simplify

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

= \frac{1}{2} sec^{2} (x) + ln ∣ cos (x) ∣

Add a constant to the solution

= \frac{1}{2} sec^{2} (x) + ln ∣ cos (x) ∣ + C

\int 5 sin^{2} (x co) s^{3} x d x

x \to 5 - lim (\frac{x - 5}{∣ x - 5 ∣})

\int sinh (x) d x

\frac{d}{d x} ((\frac{x ^{2}}{x ^{3} - 4 x})^{3})

\frac{d}{d x} (\frac{x ^{n}}{n !})

What is the integral of (tan(x))^3 ?
The integral of (tan(x))^3 is 1/2 sec^2(x)+ln|cos(x)|+C