Sum

Evaluate: `int_-π^π (1 - "x"^2) sin "x" cos^2 "x" d"x"`.

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#### Solution

`int_-π^π (1 - "x"^2) sin "x" cos^2 "x" d"x"`

We know

`int_-a^a "f" ("x")"d" "x" = 0` if f is an odd function i.e i f f (-x) = -f (x)

In the given integral,

`"f" ("x") = (1 - "x"^2) sin "x" cos^2 "x"`

⇒ `"f" (- "x") = (1- (-"x")^2) (sin (-"x")) cos^2 (-"x") = -(1 -"x"^2) sin "x" cos^2 "x"`

⇒ `"f" (-"x") = -"f" ("x")`

So, `int_-π^π (1 - "x"^2) sin "x" cos^2 "x" "dx" = 0`

Is there an error in this question or solution?

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