Advertisements
Advertisements
Question
`int_1^(sqrt3)dx/(1+x^2) ` equals:
Options
`pi/3`
`(2pi)/3`
`pi/6`
`pi/12`
Solution
`pi/12`
Explanation:
Let `I = int_1^sqrt3 dx/(1 + x^2) = [tan^-1 x]_1^sqrt3`
`= [tan^-1 sqrt3 - tan^-1 1]`
`= pi/3 - pi/4`
`= pi/12`
APPEARS IN
RELATED QUESTIONS
If `int_0^h1/(2+8x^2)dx=pi/16 `then find the value of h.
Evaluate : `∫_0^(π/2)(sin^2 x)/(sinx+cosx)dx`
Evaluate :`int_0^(pi/2)(2^(sinx))/(2^(sinx)+2^(cosx))dx`
Evaluate the definite integral:
`int_(-1)^1 (x + 1)dx`
Evaluate the definite integral:
`int_2^3 1/x dx`
Evaluate the definite integral:
`int_1^2 (4x^3 - 5x^2 + 6x + 9) dx`
Evaluate the definite integral:
`int_0^(pi/4) sin2xdx`
Evaluate the definite integral:
`int_0^(pi/2) cos 2x dx`
Evaluate the definite integral:
`int_4^5 e^x dx`
Evaluate the definite integral:
`int_0^1 dx/(1+x^2)`
Evaluate the definite integral:
`int_2^3 dx/(x^2 - 1)`
Evaluate the definite integral:
`int_0^(pi/2) cos^2 xdx`
Evaluate the definite integral:
`int_2^3 (xdx)/(x^2 + 1)`
Evaluate the definite integral:
`int_0^1 x e^(x^2) dx`
Evaluate the definite integral:
`int_1^2 (5x^2)/(x^2 + 4x + 3)`
Evaluate the definite integral:
`int_0^(pi/4) (2 sec^2 x + x^3 + 2) dx`
Evaluate the definite integral:
`int_0^pi (sin^2 x/2 - cos^2 x/2) dx`
Evaluate the definite integral:
`int_0^2 (6x +3)/(x^2 + 4)` dx
`int_0^(2/3) dx/(4+9x^2)` equals:
Evaluate : \[\int\frac{x \cos^{- 1} x}{\sqrt{1 - x^2}}dx\] .
`int_1^sqrt(3) (dx)/(1 + x^2)` equals
`int_6^(2/3) (dx)/(4 + 9x^2)` equals
Evaluate:
`int_0^π(sin^4x + cos^4x)dx`
Hence evaluate:
`int_(-2π)^(2π) (sin^4x + cos^4x)/(1 + e^x)dx`
