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

Question

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

3 Answers

Impact of this question

187004 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-04T13:08:07

I found: $sin (x) - \frac{{sin}^{3} (x)}{3} + c$ $sin(x)-(sin^3(x))/3+c$

Explanation:

Have a look:
enter image source here

Answer 2 · 2016-08-04T15:06:15

$sin (x) {cos (x)}^{2} + \frac{2}{3} {sin (x)}^{3} + C$ $sin(x)cos(x)^2+2/3sin(x)^3+C$

Explanation:

$\frac{d}{d x} (sin (x) {cos (x)}^{2}) = {cos (x)}^{3} - 2 {sin (x)}^{2} cos (x)$ $d/(dx)(sin(x)cos(x)^2)=cos(x)^3-2sin(x)^2cos(x)$

then

$\int {cos (x)}^{3} d x = sin (x) {cos (x)}^{2} + 2 \int {sin (x)}^{2} cos (x) d x$ $int cos(x)^3dx = sin(x)cos(x)^2+2intsin(x)^2cos(x)dx$

but

$\int {sin (x)}^{2} cos (x) d x = \int (\frac{1}{3} \frac{d}{d x} {sin (x)}^{3}) d x$ $intsin(x)^2cos(x)dx = int(1/3 d/(dx)sin(x)^3)dx$

finally

$\int {cos (x)}^{3} d x = sin (x) {cos (x)}^{2} + \frac{2}{3} {sin (x)}^{3} + C$ $int cos(x)^3dx = sin(x)cos(x)^2+2/3sin(x)^3+C$

Answer 3 · 2016-08-04T22:02:44

$\int {cos}^{3} x d x = sin x - \frac{1}{3} {sin}^{3} x + C$ $int cos^3 x d x=sin x-1/3sin^3 x+C$

Explanation:

$a different way...$ $"a different way..."$

$use reduction formula$ $"use reduction formula"$

$\int {cos}^{n} x d x = \frac{n - 1}{n} \int {cos}^{n - 2} x d x + \frac{{cos}^{n - 1} x \cdot sin x}{n}$ $int cos^n x d x=(n-1)/n int cos^(n-2) x d x+(cos^(n-1)x*sin x)/n$

$use n=3$ $"use n=3"$

$\int {cos}^{3} x d x = \frac{3 - 1}{3} \int {cos}^{3 - 2} x + \frac{{cos}^{3 - 1} x \cdot sin x}{3}$ $int cos^3 x d x=(3-1)/3 int cos^(3-2)x +(cos^(3-1)x*sin x)/(3)$

$\int {cos}^{3} x d x = \frac{2}{3} \int cos x d x + \frac{{cos}^{2} x \cdot sin x}{3}$ $int cos^3 x d x=2/3int cos x d x+(cos^2 x*sin x)/3$

${cos}^{2} x = 1 - {sin}^{2} x$ $cos^2x=1-sin^2x$

$\int {cos}^{3} x d x = \frac{2}{3} sin x + \frac{(1 - {sin}^{2} x) \cdot sin x}{3}$ $int cos^3 x d x=2/3 sin x+((1-sin^2 x)*sin x)/3$

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

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

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

$\int {cos}^{3} x d x = sin x - \frac{1}{3} {sin}^{3} x + C$ $int cos^3 x d x=sin x-1/3sin^3 x+C$

Science

Math

Humanities

... and beyond

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

3 Answers

Explanation:

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate ∫cos3xdxint cos^3x dx?

3 Answers

Explanation:

Explanation:

Explanation:

Related questions

Impact of this question

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