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What is integral of ∫sinx^5dx?
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| integrate of sinx^5 |
Evaluate $int sin^5{x}dx$.
Rewrite the function:
$int sin^5{x}.dx=int sinx.sin^4x.dx=int sinx(sin^2x)^2.dx=int sinx(1-cos^2x)^2.dx$
Now use u=cosx, du=-sinxdx:
$int sinx(1-cos^2x)^2.dx=int -(1-u^2)^2du=int -(1-2u^2+u^4)du$
=$-u+frac{2}{3}u^3-frac{1}{5}u^5+C$
=$-cosx+frac{2}{3}cos^3x-frac{1}{5}cos^5x+C$
Answer:
$int sin^5{x}.dx=-cosx+frac{2}{3}cos^3x-frac{1}{5}cos^5x+C$
