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I hate math help pls........how do i solve the integral from 0-pi/4 of sin(2x)sin(x)dx?
I tried to work this out by setting u = to sin(x) then du was = to cos(x).........my final answer was sqrt(2)/3 but this was not right......
2 Answers
- 1 decade agoFavorite Answer
following from the person above me,
set u=cosx then du= -sinx
then u^2(-du)
= -1/3u^3
do you mean 0 to pi/4
then -1/3(1/sqrt(2))^(3)= -1/3[2sqrt(2)]-1/3
= [-1-2sqrt(2)]/[6sqrt(2)]
- Anonymous1 decade ago
Double angle formula.....
sin(2x)=2sin(x)cos(x)
so therefore...integral sin^2(x)cos(x).....I'm sure you can take it from there