Respuesta :

Willchemistry
[tex] \int\limits8sin^4(x)dx =[/tex]

Take the constant out [tex] \int\limits a*f(x)dx=a* \int\limitsf(x)dx[/tex]

[tex]= 8 \int\limits sin^4(x)dx[/tex]

[tex] \int\limits sin^4(x)dx = - \frac{cos(x)sin^3(x)}{4} = \frac{3}{4} \int\limits sin^2(x)dx[/tex]

[tex]= 8( -\frac{cos(x)sin^3(x)}{4}+ \frac{3}{4} \int\limits sin^2(x)dx) [/tex]

Use the following identify : [tex]sin^2(x) = \frac{1-cos(2x)}{2} [/tex]

[tex]= 8(- \frac{cos(x)sin^3(x)}{4} + \frac{3}{4} \frac{1}{2}( \int\limits 1 - cos(2x)dx)[/tex]

[tex] \int\limits 1dx = x[/tex]

[tex] \int\limits cos(2x)dx = \frac{1}{2} sin(2x)[/tex]

Apply the sum rule :

[tex]=8(- \frac{cos(x)sin^3(x)}{4} + \frac{3}{4} \frac{1}{2} (x- \frac{1}{2} sin(2x))[/tex]

Simplify :

[tex]8( \frac{3}{8} (x- \frac{1}{2} sin(2x)) - \frac{1}{4} sin^3(x)cos(x))[/tex]

Therefore add  a constant  to the solution:

[tex]= 8 ( \frac{3}{8} (x- \frac{1}{2} sin(2x)) - \frac{1}{4} sin^3(x)cos(x))+C[/tex]

hope this helps!




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