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`int_(0)^(pi//2)sinxsin2xdx=?`A. `(2)/(3)`B. `(3)/(4)`C. `(5)/(6)`D. `(3)/(5)` |
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Answer» `I=(1)/(2)int_(0)^(pi//2)[cos(2x-x)-cos(2x+x)]dx=(1)/(2)int_(0)^(pi//2)(cosx-cos3x)dx` `=(1)/(2)[sinx-(sin3x)/(3)]_(0)^(pi//2)=(1)/(2)(1+(1)/(3))=(2)/(3)`. |
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