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