Advertisements
Advertisements
Question
Evaluate: `int_0^(pi/2) x sin x.dx`
Advertisements
Solution
`int_0^(pi/2) x sin x.dx`
= `[x int sinx.dx]_0^(pi/2) - int_0^(pi/2)[d/dx(x) int sin x.dx].dx`
= `[x (- cos x)]_0^(pi/2) - int_0^(pi/2) 1.(- cos x).dx`
= `-[x cosx]_0^(pi/2) + int_0^(pi/2) cosx.dx`
= `-[pi/2 cos pi/2 - 0] + [sinx]_0^(pi/2)`
= `0 + (sin pi/2 - sin 0)`
= 1
APPEARS IN
RELATED QUESTIONS
Evaluate: `int_0^(π/4) cot^2x.dx`
Evaluate: `int_0^1 (x^2 - 2)/(x^2 + 1).dx`
Evaluate: `int_0^oo xe^-x.dx`
`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` =
Evaluate: `int_(pi/6)^(pi/3) cosx "d"x`
Evaluate: `int_0^1 1/(1 + x^2) "d"x`
Evaluate: `int_0^1 "e"^x/sqrt("e"^x - 1) "d"x`
Evaluate: `int_(pi/6)^(pi/3) sin^2 x "d"x`
Evaluate: `int_0^pi cos^2 x "d"x`
Evaluate: `int_0^(pi/4) (tan^3x)/(1 + cos 2x) "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^2x)/(1 + cos x)^2 "d"x`
Evaluate: `int_0^9 sqrt(x)/(sqrt(x) + sqrt(9 - x) "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^(1/sqrt(2)) (sin^-1x)/(1 - x^2)^(3/2) "d"x`
Evaluate: `int_0^(pi/4) sec^4x "d"x`
Evaluate: `int_0^(pi/2) cos x/((1 + sinx)(2 + sinx)) "d"x`
Evaluate: `int_0^(pi/4) (sec^2x)/(3tan^2x + 4tan x + 1) "d"x`
Evaluate: `int_(1/sqrt(2))^1 (("e"^(cos^-1x))(sin^-1x))/sqrt(1 - x^2) "d"x`
Evaluate: `int_0^1 (log(x + 1))/(x^2 + 1) "d"x`
Evaluate: `int_0^(pi/4) (cos2x)/(1 + cos 2x + sin 2x) "d"x`
Evaluate: `int_0^pi 1/(3 + 2sinx + cosx) "d"x`
Evaluate: `int_0^(π/4) sec^4 x dx`
Evaluate:
`int_(π/4)^(π/2) cot^2x dx`.
Evaluate:
`int_(-π/2)^(π/2) |sinx|dx`
The value of `int_2^(π/2) sin^3x dx` = ______.
Evaluate:
`int_0^(π/2) (sin 2x)/(1 + sin^4x)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`
If `int_0^π f(sinx)dx = kint_0^π f(sinx)dx`, then find the value of k.
