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

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Answer 1 · 2016-09-04T16:45:07

$\frac{1}{192} (12 x + 3 sin 2 x - 3 sin 4 x - sin 6 x) + C$ $1/192(12x+3sin2x-3sin4x-sin6x)+C$ .

Let $I = \int {sin}^{2} x {cos}^{4} x d x$ $I=intsin^2xcos^4xdx$ .

We will use the following Identities to simplify the Integrand :-

$[1] : 2 {sin}^{2} θ = 1 - cos 2 θ, [2] : 2 {cos}^{2} θ = 1 + cos 2 θ$ $[1] :2sin^2theta=1-cos2theta, [2] : 2cos^2theta=1+cos2theta$

$[3] : 2 cos C cos D = cos (C + D) + cos (C - D)$ $[3] : 2cosCcosD=cos(C+D)+cos(C-D)$

Now, ${sin}^{2} x {cos}^{4} x = \frac{1}{8} (4 {sin}^{2} x {cos}^{2} x) (2 {cos}^{2} x)$ $sin^2xcos^4x=1/8(4sin^2xcos^2x)(2cos^2x)$

$= \frac{1}{8} {(2 sin x cos x)}^{2} (1 + cos 2 x)$ $=1/8(2sinxcosx)^2(1+cos2x)$

$= \frac{1}{8} {(sin 2 x)}^{2} (1 + cos 2 x)$ $=1/8(sin2x)^2(1+cos2x)$

$= \frac{1}{8} ({sin}^{2} 2 x) (1 + cos 2 x)$ $=1/8(sin^2 2x)(1+cos2x)$

$= \frac{1}{16} (2 {sin}^{2} 2 x) (1 + cos 2 x)$ $=1/16(2sin^2 2x)(1+cos2x)$

$= \frac{1}{16} (1 - cos 4 x) (1 + cos 2 x)$ $=1/16(1-cos4x)(1+cos2x)$

$= \frac{1}{16} (1 - cos 4 x + cos 2 x - cos 4 x cos 2 x)$ $=1/16(1-cos4x+cos2x-cos4xcos2x)$

$= \frac{1}{16} {1 - cos 4 x + cos 2 x - \frac{1}{2} (cos 6 x + cos 2 x)}$ $=1/16{1-cos4x+cos2x-1/2(cos6x+cos2x)}$

$= \frac{1}{32} (2 + cos 2 x - 2 cos 4 x - cos 6 x)$ $=1/32(2+cos2x-2cos4x-cos6x)$

$:. I=1/32int(2+cos2x-2cos4x-cos6x)dx$

$=1/32(2x+sin(2x)/2-(2sin(4x))/4-sin(6x)/6)$

$=1/192(12x+3sin2x-3sin4x-sin6x)+C$ .

Enjoy Maths.!

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