Advertisements
Advertisements
Question
`int_0^(x/4) sqrt(1 + sin 2x) "d"x` =
Options
`1/sqrt(2)`
`sqrt(2) + 1`
`2sqrt(2)`
1
Advertisements
Solution
1
APPEARS IN
RELATED QUESTIONS
Evaluate: `int_0^(π/4) cot^2x.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` = ______.
If `int_0^1 ("d"x)/(sqrt(1 + x) - sqrt(x)) = "k"/3`, then k is equal to ______.
`int_0^1 (x^2 - 2)/(x^2 + 1) "d"x` =
`int_0^4 1/sqrt(4x - x^2) "d"x` =
`int_0^(pi/2) log(tanx) "d"x` =
Evaluate: `int_0^1 |x| "d"x`
Evaluate: `int_0^1 "e"^x/sqrt("e"^x - 1) "d"x`
Evaluate: `int_0^(pi/2) (sin2x)/(1 + sin^2x) "d"x`
Evaluate: `int_(pi/6)^(pi/3) sin^2 x "d"x`
Evaluate: `int_0^(pi/2) sqrt(1 - cos 4x) "d"x`
Evaluate:
`int_0^(pi/2) cos^3x dx`
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/2) (sin^2x)/(1 + cos x)^2 "d"x`
Evaluate: `int_0^9 sqrt(x)/(sqrt(x) + sqrt(9 - x) "d"x`
Evaluate: `int_0^(pi/2) (sin^4x)/(sin^4x + cos^4x) "d"x`
Evaluate: `int_0^1 1/sqrt(3 + 2x - x^2) "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) 1/(5 + 4cos x) "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^(1/2) 1/((1 - 2x^2) sqrt(1 - x^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_(-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 1/(3 + 2sinx + cosx) "d"x`
If `int_2^e [1/logx - 1/(logx)^2].dx = a + b/log2`, then ______.
Evaluate: `int_0^1 tan^-1(x/sqrt(1 - x^2))dx`.
Evaluate:
`int_(-π/2)^(π/2) |sinx|dx`
Evaluate `int_(π/6)^(π/3) cos^2x dx`
Evaluate:
`int_-4^5 |x + 3|dx`
Evaluate:
`int_0^(π/2) (sin 2x)/(1 + sin^4x)dx`
Prove that: `int_0^1 logx/sqrt(1 - x^2)dx = π/2 log(1/2)`
