How to solve ∫ (sin^4x)(cos^2x) dx ?

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How to solve ∫ (sin^4x)(cos^2x) dx ?

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Answer 1 · 2015-09-30T07:03:04

$\int {sin}^{4} x \cdot {cos}^{2} x d x = \frac{1}{16} \cdot (x - \frac{sin (4 x)}{4} - \frac{{sin}^{3} (2 x)}{3}) + c$ $int sin^4x*cos^2x dx=1/16*(x-sin(4x)/4-(sin^3(2x))/3)+c$

Explanation:

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

$= \int (\frac{1}{2}) \cdot (1 - cos 2 x) \cdot {(\frac{sin 2 x}{2})}^{2} d x$ $=int(1/2) * (1-cos2x) * ((sin2x)/2)^2dx$

We have

$cos 2 x = 1 - 2 {sin}^{2} x$ $cos2x=1-2sin^2x$

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

and

$sin 2 x = 2 sin x \cdot cos x$ $sin2x=2sinx*cosx$

The integral becomes

$= \int (\frac{1}{2}) \cdot (1 - cos 2 x) \cdot {(\frac{sin 2 x}{2})}^{2} d x$ $=int(1/2) * (1-cos2x) * ((sin2x)/2)^2dx$

$= \frac{1}{8} (\int (1 - cos 2 x) \cdot {sin}^{2} (2 x)) d x$ $=1/8(int(1-cos2x)*sin^2(2x))dx$

$= \frac{1}{8} (\int ({sin}^{2} (2 x) - cos 2 x \cdot {sin}^{2} (2 x)) d x$ $=1/8(int(sin^2(2x)-cos2x*sin^2(2x))dx$

$= \frac{1}{8} (\int (\frac{1}{2}) \cdot 2 {sin}^{2} (2 x) d x - \int \frac{1}{2} \cdot 2 cos 2 x \cdot {sin}^{2} (2 x) d x)$ $=1/8(int(1/2)*2sin^2(2x)dx-int1/2*2cos2x*sin^2(2x)dx)$

$= \frac{1}{8} (\int (\frac{1}{2}) \cdot (1 - cos (4 x)) d x - \int \frac{1}{2} \cdot {sin}^{2} (2 x) d (sin 2 x))$ $=1/8(int(1/2)*(1-cos(4x))dx-int1/2*sin^2(2x)d(sin2x))$

$= \frac{1}{16} (\int (1 - cos (4 x)) d x - \int {sin}^{2} (2 x) d (sin 2 x))$ $=1/16(int(1-cos(4x))dx-intsin^2(2x)d(sin2x))$

$d \frac{sin 2 x}{d x} = 2 cos 2 x$ $d(sin2x)/dx=2cos2x$

$d (sin 2 x) = 2 cos 2 x \cdot d x$ $d(sin2x)=2cos2x*dx$

$= \frac{1}{16} (x - \frac{sin (4 x)}{4} - \frac{{sin}^{3} (2 x)}{3}) + c$ $=1/16(x-sin(4x)/4-sin^3(2x)/3)+c$

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How to solve ∫ (sin^4x)(cos^2x) dx ?

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