Question: "How to solve∫0πsin⁡2xsin⁡xdx"

Oxinailelpels3t14

Answered question

2022-03-31

How to solve

\int_{0}^{π} \sin 2 x \sin x d x

zevillageobau

Beginner2022-04-01Added 13 answers

There is no need for trigonometric identities, complex exponentials or the like. Observe that

\int_{0}^{π} \sin (2 x) \sin (x) d x = \int_{0}^{\frac{π}{2}} \sin (2 x) \sin (x) d x + \int_{\frac{π}{2}}^{π} \sin (2 x) \sin (x) d x

By a change of variables (

x \to π - x

) and the oddness of the integrand on the interval

[0, π]

, you find that

\int_{\frac{π}{2}}^{π} \sin (2 x) \sin (x) d x = - \int_{0}^{\frac{π}{2}} \sin (2 x) \sin (x) d x

which implies that the original integral vanishes identically.

Alannah Farmer

Beginner2022-04-02Added 11 answers

If you use the identity

\sin 2 x = 2 \cos x \sin x

then you get

\int \sin {2 x} \sin {x}, d x = 2 \int \sin^{2} {x} \cos {x}, d x

and now you can make the substitution

u = \sin x

to get

2 \int u^{2}, d u = \frac{2}{3} u^{3} + C = \frac{2}{3} \sin^{3} {x} + C

Therefore

\int_{0}^{π} \sin {2 x} \sin {x}, d x = {(\frac{2}{3} \sin^{3} {x})}_{0}^{π} = 0

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