Advertisements
Advertisements
प्रश्न
Evaluate: `int_0^(pi/2) (sin2x)/(1 + sin^2x) "d"x`
Advertisements
उत्तर
`int_0^(pi/2) (sin2x)/(1 + sin^2x) "d"x = [log|1 + sin^2x|]_0^(pi/2)` .......`[∵ int ("f'"(x))/("f"(x)) "d"x = log|"f"(x)| + "c"]`
= `log |1 + sin^2(pi/2)| - log|1 + sin^2 0|`
= log |1 + 1| – log 1
= log 2 – 0
= log 2
APPEARS IN
संबंधित प्रश्न
Evaluate: `int_0^(π/4) cot^2x.dx`
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`
Choose the correct option from the given alternatives :
`int_0^(pi/2) (sin^2x*dx)/(1 + cosx)^2` = ______.
`int_(pi/5)^((3pi)/10) sinx/(sinx + cosx) "d"x` =
Let I1 = `int_"e"^("e"^2) 1/logx "d"x` and I2 = `int_1^2 ("e"^x)/x "d"x` then
`int_0^4 1/sqrt(4x - x^2) "d"x` =
Evaluate: `int_(pi/6)^(pi/3) cosx "d"x`
Evaluate: `int_0^1(x + 1)^2 "d"x`
Evaluate: `int_(pi/6)^(pi/3) sin^2 x "d"x`
Evaluate:
`int_0^(pi/2) cos^3x dx`
Evaluate: `int_0^pi cos^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_3^8 (11 - x)^2/(x^2 + (11 - x)^2) "d"x`
Evaluate: `int_(-1)^1 |5x - 3| "d"x`
Evaluate: `int_(-4)^2 1/(x^2 + 4x + 13) "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_(-1)^1 1/("a"^2"e"^x + "b"^2"e"^(-x)) "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/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 x*sinx*cos^2x* "d"x`
Evaluate: `int_(-1)^1 (1 + x^2)/(9 - x^2) "d"x`
Evaluate: `int_0^(pi/4) (cos2x)/(1 + cos 2x + sin 2x) "d"x`
Evaluate: `int_0^(pi/4) log(1 + tanx) "d"x`
Evaluate: `int_0^(π/4) sec^4 x dx`
Evaluate:
`int_(π/4)^(π/2) cot^2x dx`.
Evaluate:
`int_0^(π/2) sin^8x dx`
Evaluate `int_(π/6)^(π/3) cos^2x dx`
The value of `int_2^(π/2) sin^3x dx` = ______.
Find the value of ‘a’ if `int_2^a (x + 1)dx = 7/2`
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)`
If `int_0^π f(sinx)dx = kint_0^π f(sinx)dx`, then find the value of k.
