Advertisements
Advertisements
Question
Evaluate: `int_0^(pi/2) 1/(5 + 4cos x) "d"x`
Solution
Let I = `int_0^(pi/2) 1/(5 + 4cos x) "d"x`
Put `tan (x/2)` = t
∴ x = 2tan−1t
∴ dx = `2/(1 + "t"^2)` dt and cos x = `(1 - "t"^2)/(1 + "t"^2)`
When x = 0, t = 0 and when x = `pi/2`, t = 1
∴ I = `int_0^1 1/(5 + 4((1 - "t"^2)/(1 + "t"^2))) xx 2/(1 + "t"^2) "dt"`
= `2int_0^1 1/(5 + 5"t" + 4 - 4"t"^2) "dt"`
= `2int_0^1 1/(9 + "t"^2) "dt"`
= `2int_0^1 1/("t"^2 + 3^2) "dt"`
= `2[1/3 tan^-1("t"/3)]_0^1`
= `2/3[tan^-1(1/3) - tan^-1(0)]`
= `2/3 tan^-1 (1/3)`
APPEARS IN
RELATED QUESTIONS
Evaluate: `int_0^oo xe^-x.dx`
Evaluate the following:
`int_0^a (1)/(x + sqrt(a^2 - x^2)).dx`
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` =
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` =
`int_0^(pi/2) log(tanx) "d"x` =
Evaluate: `int_0^1 1/(1 + x^2) "d"x`
Evaluate: `int_0^(pi/4) sec^2 x "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_0^1(x + 1)^2 "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/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_(-1)^1 |5x - 3| "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_0^1 "t"^2 sqrt(1 - "t") "dt"`
Evaluate: `int_(1/sqrt(2))^1 (("e"^(cos^-1x))(sin^-1x))/sqrt(1 - x^2) "d"x`
Evaluate: `int_0^pi x*sinx*cos^2x* "d"x`
Evaluate: `int_0^(pi/4) log(1 + tanx) "d"x`
`int_0^(π/2) sin^6x cos^2x.dx` = ______.
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_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.
