Evaluate the integral integrate of (sin(x))^2(cos(x))^4 dx

Saturday, November 6, 2010

Evaluate the integral integrate of (sin(x))^2(cos(x))^4 dx

$\displaystyle{\sin{{2}}}{x}=$
2sinx*cosx

$\displaystyle{{\left({\sin{{2}}}{x}\right)}}^{{2}}={4}{{\left({\sin{{x}}}\right)}}^{{2}}\cdot{{\left({\cos{{\left({x}\right)}}}\right)}}^{{2}}$

$\displaystyle{{\left({\sin{{x}}}\right)}}^{{2}}\cdot{{\left({\cos{{\left({x}\right)}}}\right)}}^{{2}}=\frac{{1}}{{4}}{{\left({\sin{{2}}}{x}\right)}}^{{2}}$

$\displaystyle{\cos{{2}}}{A}=$
2cos^2A-1 = 1-2sin^A

$\displaystyle{{\cos}}^{{2}}{A}=\frac{{1}}{{2}}{\left({\cos{{2}}}{A}+{1}\right)}$

$\displaystyle{{\sin}}^{{2}}{A}=$
1/2(1-cos2A)

int(sin(x))^2(cos(x))^4 dx

$\displaystyle=\int{{\left({\sin{{x}}}\right)}}^{{2}}\cdot{{\left({\cos{{x}}}\right)}}^{{2}}\cdot{{\left({\cos{{x}}}\right)}}^{{2}}{\left.{d}{x}\right.}$

$\displaystyle=\int$
(1/4(sin2x)^2)(1/2(cos2x+1))dx

$\displaystyle=\frac{{1}}{{8}}\int{\left[{{\left({\sin{{2}}}{x}\right)}}^{{2}}{\cos{{2}}}{x}+{{\left({\sin{{2}}}{x}\right)}}^{{2}}\right]}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{1}}{{8}}{\left[\int{{\left({\sin{{2}}}{x}\right)}}^{{2}}{\cos{{2}}}{x}{\left.{d}{x}\right.}+\int\frac{{1}}{{2}}{\left({1}-{\cos{{4}}}{x}\right)}{\left.{d}{x}\right.}\right]}$

$\displaystyle\int{{\left({\sin{{2}}}{x}\right)}}^{{2}}{\cos{{2}}}{x}{\left.{d}{x}\right.}$

Let
$\displaystyle{t}={\sin{{2}}}{x}$

$\displaystyle{\left.{d}{t}\right.}={2}{\cos{{2}}}{x}{\left.{d}{x}\right.}$

dt/2 = cos2xdx

$\displaystyle\int{{\left({\sin{{2}}}{x}\right)}}^{{2}}{\cos{{2}}}{x}{\left.{d}{x}\right.}$

$\displaystyle=\int{{t}}^{{2}}\frac{{\left.{d}{t}}}{{2}}$

= 1/2(t^3/3)

$\displaystyle=\frac{{{t}}^{{3}}}{{6}}$

$\displaystyle=$
(sin2x)^3/6

$\displaystyle\int\frac{{1}}{{2}}{\left({1}-{\cos{{4}}}{x}\right)}{\left.{d}{x}\right.}$

$\displaystyle=\frac{{1}}{{2}}{\left[\int{1}{\left.{d}{x}\right.}-\int{\cos{{4}}}{x}{\left.{d}{x}\right.}\right]}$

$\displaystyle=\frac{{1}}{{2}}{\left({x}-\frac{{1}}{{4}}{\left({\sin{{4}}}{x}\right)}\right)}$

$\displaystyle=\frac{{1}}{{8}}{\left({4}{x}-{\sin{{4}}}{x}\right)}$

$\displaystyle\int{{\left({\sin{{\left({x}\right)}}}\right)}}^{{2}}{{\left({\cos{{\left({x}\right)}}}\right)}}^{{4}}{\left.{d}{x}\right.}$

$\displaystyle=$
1/8[int(sin2x)^2cos2xdx+int(sin2x)^2cos2xdx]

$\displaystyle=$
1/8[(sin2x)^3/6+1/8(4x-sin4x)]+C

$\displaystyle=\frac{{1}}{{24}}{\left[{4}{{\left({\sin{{2}}}{x}\right)}}^{{3}}+{3}{\left({4}{x}-{\sin{{4}}}{x}\right)}\right]}+{C}$

C is a constant since this is indefinite integral.

$\displaystyle\int{{\left({\sin{{\left({x}\right)}}}\right)}}^{{2}}{{\left({\cos{{\left({x}\right)}}}\right)}}^{{4}}{\left.{d}{x}\right.}=\frac{{1}}{{24}}{\left[{4}{{\left({\sin{{2}}}{x}\right)}}^{{3}}+{3}{\left({4}{x}-{\sin{{4}}}{x}\right)}\right]}+{C}$

Spooky Notes

Saturday, November 6, 2010

Evaluate the integral integrate of (sin(x))^2(cos(x))^4 dx

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