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zavuch27 [327]
3 years ago
15

Intergal of 8sin^4(x)dx

Mathematics
1 answer:
Romashka-Z-Leto [24]3 years ago
3 0
\int\limits8sin^4(x)dx =

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

= 8 \int\limits sin^4(x)dx

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

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


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

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

\int\limits 1dx = x

\int\limits cos(2x)dx =  \frac{1}{2} sin(2x)

Apply the sum rule :

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

Simplify :

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

Therefore add  a constant  to the solution:

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

hope this helps!




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