Advertisements
Advertisements
प्रश्न
Evaluate: `int_0^pi x*sinx*cos^2x* "d"x`
Advertisements
उत्तर
Let I = `int_0^pi x*sinx*cos^2x* "d"x` ......(i)
∴ I = `int_0^pi (pi - x)*sin(pi - x)*[cos(pi - x)]^2 "d"x` ......`[∵ int_0^"a" "f"(x) "d"x = int_0^"a" "f"("a" - x) "d"x]`
∴ I = `int_0^pi (pi - x)*sinx( - cos x)^2 "d"x`
∴ I = `int_0^pi (pi - x)* sinx cos^2x "d"x` .....(ii)
Adding (i) and (ii), we get
2I = `int_0^pi x* sinx * cos^2x "d"x + int_0^pi (pi - x) * sinx cos^2x "d"x`
= `int_0pi (x + pi - x)* sinx cos^2x "d"x`
∴ 2I = `pi int_0^pi sinx cos^2x "d"x`
Put cos x = t
∴ − sin x dx = dt
∴ sin x dx = − dt
When x = 0, t = 1 and when x = π, t = −1
∴ 2I = `pi int_1^(-1) "t"^2 (- "dt")`
∴ I = `pi/2 int_(-1)^1 "t"^2 "dt"` .......`[∵ int_"a"^"b" "f"(x) "d"x = -int_"b"^"a" "f"(x) "d"x]`
= `pi/2 xx 2 int_0^1 "t"^2 "dt"` ......[∵ t2 is an even function]
= `pi ["t"^2/3]_0^1`
= `pi/3(1^3 - 0)`
∴ I = `pi/3`
APPEARS IN
संबंधित प्रश्न
Evaluate: `int_0^1 (x^2 - 2)/(x^2 + 1).dx`
Evaluate: `int_0^(pi/2) x sin x.dx`
Evaluate: `int_0^oo xe^-x.dx`
Evaluate the following:
`int_0^a (1)/(x + sqrt(a^2 - x^2)).dx`
Choose the correct option from the given alternatives :
`int_0^(pi/2) (sin^2x*dx)/(1 + cosx)^2` = ______.
`int_0^(x/4) sqrt(1 + sin 2x) "d"x` =
If `int_0^1 ("d"x)/(sqrt(1 + x) - sqrt(x)) = "k"/3`, then k is equal to ______.
`int_(pi/5)^((3pi)/10) sinx/(sinx + cosx) "d"x` =
`int_0^1 (x^2 - 2)/(x^2 + 1) "d"x` =
Let I1 = `int_"e"^("e"^2) 1/logx "d"x` and I2 = `int_1^2 ("e"^x)/x "d"x` then
Evaluate: `int_(- pi/4)^(pi/4) x^3 sin^4x "d"x`
Evaluate: `int_0^1 1/sqrt(1 - x^2) "d"x`
Evaluate: `int_1^2 x/(1 + x^2) "d"x`
Evaluate: `int_0^(pi/2) (sin2x)/(1 + sin^2x) "d"x`
Evaluate: `int_0^(pi/2) sqrt(1 - cos 4x) "d"x`
Evaluate: `int_0^pi cos^2 x "d"x`
Evaluate: `int_0^(pi/4) cosx/(4 - sin^2 x) "d"x`
Evaluate: `int_1^3 (cos(logx))/x "d"x`
Evaluate: `int_0^(pi/2) (sin^4x)/(sin^4x + cos^4x) "d"x`
Evaluate: `int_3^8 (11 - x)^2/(x^2 + (11 - x)^2) "d"x`
Evaluate: `int_(-4)^2 1/(x^2 + 4x + 13) "d"x`
Evaluate: `int_0^1 x* tan^-1x "d"x`
Evaluate: `int_0^(pi/2) cos x/((1 + sinx)(2 + sinx)) "d"x`
Evaluate: `int_0^3 x^2 (3 - x)^(5/2) "d"x`
Evaluate: `int_0^(pi/4) (sec^2x)/(3tan^2x + 4tan x + 1) "d"x`
Evaluate: `int_(-1)^1 (1 + x^2)/(9 - x^2) "d"x`
Evaluate: `int_0^(pi/4) log(1 + tanx) "d"x`
`int_0^(π/2) sin^6x cos^2x.dx` = ______.
Evaluate:
`int_(π/4)^(π/2) cot^2x dx`.
Evaluate: `int_0^1 tan^-1(x/sqrt(1 - x^2))dx`.
Evaluate:
`int_0^(π/2) sin^8x dx`
Evaluate:
`int_-4^5 |x + 3|dx`
Evaluate:
`int_0^(π/2) sinx/(1 + cosx)^3 dx`
Prove that: `int_0^1 logx/sqrt(1 - x^2)dx = π/2 log(1/2)`
Evaluate `int_(-π/2)^(π/2) sinx/(1 + cos^2x)dx`
