How do you evaluate the integral $\int {tan}^{3} θ$ $int tan^3theta$ ?

Question

How do you evaluate the integral $\int {tan}^{3} θ$ $int tan^3theta$ ?

1 Answer

Impact of this question

2112 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-21T02:08:58

$\int {tan}^{3} x d x = \frac{1}{2} {tan}^{2} x + ln | cos x | + C$ $int tan^3x dx = 1/2tan^2x +ln abs(cosx) + C$

Explanation:

Use the trigonometric identity:

${tan}^{2} x = {sec}^{2} x - 1$ $tan^2x = sec^2x-1$

to get:

$\int {tan}^{3} x d x = \int ({sec}^{2} x - 1) tan x d x = \int tan x {sec}^{2} x d x - \int tan x d x$ $int tan^3x dx = int (sec^2x -1) tanx dx= int tanxsec^2xdx -int tanx dx$

Solve the first integral using: $d (tan x) = {sec}^{2} x d x$ $d(tanx) = sec^2xdx$

$\int tan x {sec}^{2} d x = \int tan x d (tan x) = \frac{1}{2} {tan}^{2} x + C_{1}$ $int tanx sec^2 dx = int tanx d(tanx) = 1/2tan^2x +C_1$

For the second integral:

$\int tan x d x = \int \frac{sin x}{cos x} d x = - \int \frac{d (cos x)}{cos x} = - ln | cos x | + C_{2}$ $int tanx dx = int sinx/cosx dx = - int (d(cosx))/cosx = -ln abs(cosx) + C_2$

Putting it together:

$\int {tan}^{3} x d x = \frac{1}{2} {tan}^{2} x + ln | cos x | + C$ $int tan^3x dx = 1/2tan^2x +ln abs(cosx) + C$

Science

Math

Humanities

... and beyond

How do you evaluate the integral $\int {tan}^{3} θ$ $int tan^3theta$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you evaluate the integral int tan^3theta∫tan3θint tan^3theta?

1 Answer

Explanation:

Related questions

Impact of this question

How do you evaluate the integral $\int {tan}^{3} θ$ $int tan^3theta$ ?