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Lösung 2.3:1b

Aus Online Mathematik Brückenkurs 2

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Version vom 10:26, 11. Mär. 2009

If we look at the formula for integration by parts,

f(x)g(x)dx=F(x)g(x)−F(x)g(x)dx 

we see that if we choose f(x)=sinx and g(x)=x+1, then the factor g(x) will be differentiated to a constant on the right-hand side of the integral. Naturally, this presupposes that we can find a primitive function for f(x) (which we can) and that we can then integrate it. Let's try!

(x+1)sinxdx=(x+1)(−cosx)−1(−cosx)dx=−(x+1)cosx+cosxdx=−(x+1)cosx+sinx+C.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.3:1b“