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
find `∫_2^4 x/(x^2 + 1)dx`
Evaluate :
`int_e^(e^2) dx/(xlogx)`
Evaluate the integral by using substitution.
`int_0^2 xsqrt(x+2)` (Put x + 2 = `t^2`)
Evaluate the integral by using substitution.
`int_1^2 (1/x- 1/(2x^2))e^(2x) dx`
The value of the integral `int_(1/3)^4 ((x- x^3)^(1/3))/x^4` dx is ______.
If `f(x) = int_0^pi t sin t dt`, then f' (x) is ______.
`int 1/(1 + cos x)` dx = _____
A) `tan(x/2) + c`
B) `2 tan (x/2) + c`
C) -`cot (x/2) + c`
D) -2 `cot (x/2)` + c
Evaluate of the following integral:
(i) \[\int x^4 dx\]
Evaluate of the following integral:
Evaluate:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate each of the following integral:
Evaluate the following integral:
Evaluate
\[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\]
Evaluate the following integral:
Find : \[\int\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\] .
Evaluate: \[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x}dx\] .
Evaluate: `int_ e^x ((2+sin2x))/cos^2 x dx`
If `I_n = int_0^(pi/4) tan^n theta "d"theta " then " I_8 + I_6` equals ______.
`int_0^(pi4) sec^4x "d"x` = ______.
`int_0^1 sin^-1 ((2x)/(1 + x^2))"d"x` = ______.
Evaluate the following:
`int ("e"^(6logx) - "e"^(5logx))/("e"^(4logx) - "e"^(3logx)) "d"x`
Evaluate the following:
`int "dt"/sqrt(3"t" - 2"t"^2)`
Find: `int (dx)/sqrt(3 - 2x - x^2)`
Evaluate: `int_0^(π/2) sin 2x tan^-1 (sin x) dx`.
If \[t=\tan^{-1}x\], what is \[dt\]?
What is the value of \[\int_{0}^{1}\frac{\tan^{-1}x}{1+x^2}\,dx\]?
