How do you calculate this ? #intsin(x)cos(x)dx#

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Answer 1 · 2017-09-26T07:56:35

#=-1/4cos2x+C#

#intsinxcosxdx#

two ways are shown below and a third suggested

(1) we can use the trig identity

#sin2x=2sinxcosx#

#intsinxcosxdx=int1/2sin2xdx#

#=1/2intsin2xdx#

#=-1/4cos2x+C#

(2) we can do it by inspection

recognising

#d/(dx)(sin^2x)=2sinxcosx#

by the chain rule

#=>intsinxcosxdx=1/2sin^2x+C#

which can be shown ot be equivalent to the previous solution by the use of trig identities. This is left for the reader as an exercise

(3) substitution

let #u=sinx " or " u=cosx#

this also is left for the reader as an exercise