Advertisements
Advertisements
Question
`int_0^(pi/2) log(tanx) "d"x` =
Options
`pi/8(log2)`
0
`- pi/8 (log2)`
`pi/2 (log2)`
Advertisements
Solution
0
RELATED QUESTIONS
Evaluate: `int_0^1 (x^2 - 2)/(x^2 + 1).dx`
Evaluate: `int_0^(pi/2) x sin x.dx`
Evaluate: `int_0^π sin^3x (1 + 2cosx)(1 + cosx)^2.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_(pi/5)^((3pi)/10) sinx/(sinx + cosx) "d"x` =
`int_0^4 1/sqrt(4x - x^2) "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_(pi/6)^(pi/3) sin^2 x "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_0^9 sqrt(x)/(sqrt(x) + sqrt(9 - x) "d"x`
Evaluate: `int_0^(pi/2) (sin^4x)/(sin^4x + cos^4x) "d"x`
Evaluate: `int_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_0^1 1/sqrt(3 + 2x - x^2) "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) 1/(5 + 4cos x) "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^"a" 1/(x + sqrt("a"^2 - x^2)) "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_0^1 (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/4) log(1 + tanx) "d"x`
Evaluate: `int_0^pi 1/(3 + 2sinx + cosx) "d"x`
Evaluate: `int_0^(π/4) sec^4 x dx`
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_0^(π/2) sin^8x dx`
Evaluate:
`int_(π/6)^(π/3) (root(3)(sinx))/(root(3)(sinx) + root(3)(cosx))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.
