What is the integral of $\int {sin}^{2} (x) {cos}^{4} (x) d x$ $int sin^2(x)cos^4(x) dx$ ?

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What is the integral of $\int {sin}^{2} (x) {cos}^{4} (x) d x$ $int sin^2(x)cos^4(x) dx$ ?

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Answer 1 · 2016-06-05T06:43:57

$\frac{1}{16} (x - \frac{1}{4} sin (4 x)) + \frac{1}{6} {sin}^{3} (x) {cos}^{3} (x) + C$ $\frac{1}{16}(x-\frac{1}{4}\sin (4x))+\frac{1}{6}\sin ^3(x)\cos ^3(x)+C$

Explanation:

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

Applying integral reduction,

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

so,
$\int {sin}^{2} (x) {cos}^{4} (x) d x$ $\int \sin ^2(x)\cos ^4(x)dx$
$= \frac{{sin}^{3} (x) {cos}^{3} (x)}{6} + \frac{3}{6} \int {cos}^{2} (x) {sin}^{2} (x) d x$ $=\frac{\sin ^3(x)\cos ^3(x)}{6}+\frac{3}{6}\int \cos ^2(x)\sin ^2(x)dx$

$= \frac{{sin}^{3} (x) {cos}^{3} (x)}{6} + \frac{3}{6} \int {cos}^{2} (x) {sin}^{2} (x) d x$ $=\frac{\sin ^3(x)\cos ^3(x)}{6}+\frac{3}{6}\int \cos ^2(x)\sin ^2(x)dx$

We know,
$\int {cos}^{2} (x) {sin}^{2} (x) d x = \frac{1}{8} (x - \frac{1}{4} sin (4 x))$ $\int \cos ^2(x)\sin ^2(x)dx=\frac{1}{8}(x-\frac{1}{4}\sin (4x))$

Then,
$= \frac{{sin}^{3} (x) {cos}^{3} (x)}{6} + \frac{3}{6} \frac{1}{8} (x - \frac{1}{4} sin (4 x))$ $=\frac{\sin ^3(x)\cos ^3(x)}{6}+\frac{3}{6}\frac{1}{8}(x-\frac{1}{4}\sin (4x))$

Simplifying,
$= \frac{1}{16} (x - \frac{1}{4} sin (4 x)) + \frac{1}{6} {sin}^{3} (x) {cos}^{3} (x)$ $=\frac{1}{16}(x-\frac{1}{4}\sin (4x))+\frac{1}{6}\sin ^3(x)\cos ^3(x)$

Adding constant to the solution,
$= \frac{1}{16} (x - \frac{1}{4} sin (4 x)) + \frac{1}{6} {sin}^{3} (x) {cos}^{3} (x) + C$ $=\frac{1}{16}(x-\frac{1}{4}\sin (4x))+\frac{1}{6}\sin ^3(x)\cos ^3(x)+C$

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