What is the integral of (tan^3(x))/(sec^2(x)-1) ?

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

Popular >

integral of (tan^3(x))/(sec^2(x)-1)

\int \frac{tan ^{3} ( x )}{sec ^{2} ( x ) - 1} d x

- ln ∣ cos (x) ∣ + C

Solution steps

\int \frac{tan ^{3} ( x )}{sec ^{2} ( x ) - 1} d x

Simplify

\frac{tan ^{3} ( x )}{sec ^{2} ( x ) - 1} : tan (x)

= \int tan (x) d x

Rewrite using trig identities

= \int \frac{sin ( x )}{cos ( x )} d x

Apply u-substitution

: \int - \frac{1}{u} d u

= \int - \frac{1}{u} d u

Take the constant out:

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

= - \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = cos (x)

= - ln ∣ cos (x) ∣

Add a constant to the solution

= - ln ∣ cos (x) ∣ + C

y = ln (x ln (x))

\int_{0}^{9} \frac{1}{81 + x ^{2}} d x

\int sec^{5} (t) d t

\frac{\partial ^{2}}{\partial x ^{2}} (x^{2} - e^{y^{2}})

ln ∣ sec (θ) + tan (θ) ∣

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