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Question #e1ae5

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Answer 1 · 2016-12-10T17:04:05

$\int_{0}^{\frac{π}{4}} 16 {sin}^{4} (x) {cos}^{2} (x) d x = \frac{π}{4} - \frac{1}{3}$ $int_0^(pi/4)16sin^4(x)cos^2(x)dx=pi/4-1/3$

Explanation:

First, let's find the indefinite integral $\int 16 {sin}^{4} (x) {cos}^{2} (x) d x$ $int16sin^4(x)cos^2(x)dx$ . To do so, we will make use of the formulas

${sin}^{2} (x) = \frac{1 - cos (2 x)}{2}$ $sin^2(x) = (1-cos(2x))/2$
$sin (2 x) = 2 sin (x) cos (x)$ $sin(2x) = 2sin(x)cos(x)$

$\int 16 {sin}^{4} (x) {cos}^{2} (x) d x = 16 \int {sin}^{2} (x) {(sin (x) cos (x))}^{2} d x$ $int16sin^4(x)cos^2(x)dx = 16intsin^2(x)(sin(x)cos(x))^2dx$

$= 16 \int {sin}^{2} (x) {(\frac{1}{2} sin (2 x))}^{2} d x$ $=16intsin^2(x)(1/2sin(2x))^2dx$

$= 4 \int {sin}^{2} (x) {sin}^{2} (2 x) d x$ $=4intsin^2(x)sin^2(2x)dx$

$= 4 \int \frac{1 - cos (2 x)}{2} {sin}^{2} (2 x) d x$ $=4int(1-cos(2x))/2sin^2(2x)dx$

$= 2 \int ({sin}^{2} (2 x) - {sin}^{2} (2 x) cos (2 x)) d x$ $=2int(sin^2(2x)-sin^2(2x)cos(2x))dx$

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

Let's evaluate these integrals separately.

$2 \int {sin}^{2} (2 x) d x = 2 \int \frac{1 - cos (4 x)}{2} d x$ $2intsin^2(2x)dx = 2int(1-cos(4x))/2dx$

$= \int d x - \int cos (4 x) d x$ $=intdx - intcos(4x)dx$

$= x - \frac{sin (4 x)}{4} + C$ $=x-sin(4x)/4+C$

For the second integral, we make the substitution

$u = sin (2 x) \Rightarrow d u = 2 cos (2 x) d x$ $u = sin(2x) => du = 2cos(2x)dx$

Then

$2 \int {sin}^{2} (2 x) cos (2 x) d x = \int u^{2} d u$ $2intsin^2(2x)cos(2x)dx = intu^2du$

$= \frac{u^{3}}{3} + C$ $=u^3/3+C$

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

Putting the above together, we get

$\int 16 {sin}^{4} (x) {cos}^{2} (x) d x$ $int16sin^4(x)cos^2(x)dx$

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

$= x - \frac{sin (4 x)}{4} - \frac{{sin}^{3} (2 x)}{3} + C$ $=x-sin(4x)/4-sin^3(2x)/3+C$

We can now evaluate the definite integral.

$\int_{0}^{\frac{π}{4}} 16 {sin}^{4} (x) {cos}^{2} (x) d x = {[x - \frac{sin (4 x)}{4} - \frac{{sin}^{3} (2 x)}{3}]}_{0}^{\frac{π}{4}}$ $int_0^(pi/4)16sin^4(x)cos^2(x)dx = [x-sin(4x)/4-sin^3(2x)/3]_0^(pi/4)$

$= (\frac{π}{4} - \frac{sin (π)}{4} - \frac{{sin}^{3} (\frac{π}{2})}{3})$ $=(pi/4-sin(pi)/4-sin^3(pi/2)/3)$

$- (0 - \frac{sin (0)}{4} - \frac{{sin}^{3} (0)}{3})$ $-(0-sin(0)/4-sin^3(0)/3)$

$= \frac{π}{4} - \frac{0}{4} - \frac{1}{3} - 0 + 0 + 0$ $=pi/4-0/4-1/3-0+0+0$

$= \frac{π}{4} - \frac{1}{3}$ $=pi/4-1/3$

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