How do you integrate $\int {cos}^{2} (x) {tan}^{3} (x) d x$ $int cos^2(x) tan^3(x) dx$ ?

Question

How do you integrate $\int {cos}^{2} (x) {tan}^{3} (x) d x$ $int cos^2(x) tan^3(x) dx$ ?

1 Answer

Impact of this question

15487 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-23T07:08:15

I got:

$\frac{{cos}^{2} x}{2} - ln | cos x | + C$ $cos^2x/2 - ln|cosx| + C$

and Wolfram Alpha agrees.

Note that ${tan}^{3} x = \frac{{sin}^{3} x}{{cos}^{3} x}$ $tan^3x = sin^3x/cos^3x$ . Therefore:

$\int {cos}^{2} x {tan}^{3} x d x$ $color(blue)(int cos^2xtan^3xdx)$

$= \int \frac{{sin}^{3} x}{cos x} d x$ $= int (sin^3x)/cosxdx$

Then, you can use $u$ $u$ -substitution. When $u = cos x$ $u = cosx$ , $d u = - sin x d x$ $du = -sinxdx$ . To do that, you need to turn ${sin}^{2} x$ $sin^2x$ into $1 - {cos}^{2} x$ $1-cos^2x$ .

$= \int \frac{(1 - {cos}^{2} x) sin x}{cos x} d x$ $= int ((1-cos^2x)sinx)/(cosx)dx$

Hence:

$= - \int \frac{(1 - {cos}^{2} x) (- sin x)}{cos x} d x$ $= -int ((1-cos^2x)(-sinx))/(cosx)dx$

$= - \int \frac{1 - u^{2}}{u} d u$ $= -int (1-u^2)/(u)du$

$= \int \frac{u^{2} - 1}{u} d u$ $= int (u^2 - 1)/(u)du$

$= \int u - \frac{1}{u} d u$ $= int u - 1/udu$

$= \frac{u^{2}}{2} - ln | u |$ $= u^2/2 - ln|u|$

Re-substitute to get:

$\Rightarrow \frac{{cos}^{2} x}{2} - ln | cos x | + C$ $=> color(blue)(cos^2x/2 - ln|cosx| + C)$

Science

Math

Humanities

... and beyond

How do you integrate $\int {cos}^{2} (x) {tan}^{3} (x) d x$ $int cos^2(x) tan^3(x) dx$ ?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate ∫cos2(x)tan3(x)dxint cos^2(x) tan^3(x) dx?

1 Answer

Related questions

Impact of this question

How do you integrate $\int {cos}^{2} (x) {tan}^{3} (x) d x$ $int cos^2(x) tan^3(x) dx$ ?