How to solve \(\displaystyle{\int_{{0}}^{\pi}}{\sin{{2}}}{x}{\sin{{x}}}{\left.{d}{x}\right.}\)

Oxinailelpels3t14

Oxinailelpels3t14

Answered question

2022-03-31

How to solve
0πsin2xsinxdx

Answer & Explanation

zevillageobau

zevillageobau

Beginner2022-04-01Added 13 answers

There is no need for trigonometric identities, complex exponentials or the like. Observe that
0πsin(2x)sin(x)dx=0π2sin(2x)sin(x)dx+π2πsin(2x)sin(x)dx
By a change of variables (xπx) and the oddness of the integrand on the interval [0,π], you find that
π2πsin(2x)sin(x)dx=0π2sin(2x)sin(x)dx
which implies that the original integral vanishes identically.
Alannah Farmer

Alannah Farmer

Beginner2022-04-02Added 11 answers

If you use the identity sin2x=2cosxsinx then you get
sin{2x}sin{x},dx=2sin2{x}cos{x},dx
and now you can make the substitution u=sinx to get
2u2,du=23u3+C=23sin3{x}+C
Therefore
0πsin{2x}sin{x},dx=(23sin3{x})0π=0

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