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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Use the basic trigonometric identity:

\frac{1}{sec ( x )} = cos (x)

= \int tan^{3} (x) cos^{2} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= - ln ∣ u ∣ + \frac{u ^{2}}{2}

Substitute back

u = cos (x)

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

Add a constant to the solution

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

\int (x - 1) (x - 2) d x

\int \frac{1}{9 - ( x + 2 ) ^{2}} d x

\int t^{2} + 2 t d t

\int 3 x x - 8 d x

\int sin (m x) cos (n x) d x

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