Advertisements
Advertisements
प्रश्न
Evaluate: `int_-π^π (1 - "x"^2) sin "x" cos^2 "x" d"x"`.
Advertisements
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`
APPEARS IN
संबंधित प्रश्न
Evaluate: `int (1+logx)/(x(2+logx)(3+logx))dx`
Evaluate: `int1/(xlogxlog(logx))dx`
Evaluate : `int1/(3+5cosx)dx`
Evaluate: `intsinsqrtx/sqrtxdx`
Evaluate of the following integral:
Evaluate of the following integral:
Evaluate:
Evaluate:
Evaluate the following definite integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate each of the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate: `int_-1^2 (|"x"|)/"x"d"x"`.
If `I_n = int_0^(pi/4) tan^n theta "d"theta " then " I_8 + I_6` equals ______.
Find: `int (dx)/sqrt(3 - 2x - x^2)`
`int_0^1 x^2e^x dx` = ______.
The value of `int_0^1 (x^4(1 - x)^4)/(1 + x^2) dx` is
Evaluate: `int x/(x^2 + 1)"d"x`
