Advertisements
Advertisements
Question
Evaluate: `int_0^1 x* tan^-1x "d"x`
Advertisements
Solution
Let I = `int_0^1 x tan^-1x "d"x`
= `[tan^-1 x int x "d"x]_0^1 - int_0^1["d"/("d"x)(tan^-1x) int x "d"x]"d"x`
= `[tan^-1x* x^2/2]_0^1 -int_0^1 1/(1 + x^2)*x^2/2 "d"x`
= `[x^2/2 tan^-1x]_0^1 - 1/2 int_0^1 x^2/(1 + x^2) "d"x`
= `[1/2 tan^-1 - 0] - 1/2 int (x^2 + 1 - 1)/(1 + x^2) "d"x`
= `1/2* pi/4 - 1/2 int_0^1 (1 - 1/(1 + x^2)) "d"x`
= `pi/8 - 1/2[x - tan^-1x]_0^1`
= `pi/8 - 1/2[(1 - tan^-1 1) - (0 - tan^-1 0)]`
= `pi/8 - 1/2(1 - pi/4 - 0)`
= `pi/8 - 1/2 + pi/8`
∴ I = `pi/4 - 1/2`
APPEARS IN
RELATED QUESTIONS
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`
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` = ______.
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` =
`int_0^1 (x^2 - 2)/(x^2 + 1) "d"x` =
Let I1 = `int_"e"^("e"^2) 1/logx "d"x` and I2 = `int_1^2 ("e"^x)/x "d"x` then
Evaluate: `int_(pi/6)^(pi/3) cosx "d"x`
Evaluate: `int_(- pi/4)^(pi/4) x^3 sin^4x "d"x`
Evaluate: `int_0^(pi/2) (sin2x)/(1 + sin^2x) "d"x`
Evaluate: `int_0^1(x + 1)^2 "d"x`
Evaluate: `int_0^(pi/4) cosx/(4 - sin^2 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_0^(1/sqrt(2)) (sin^-1x)/(1 - x^2)^(3/2) "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_0^1 "t"^2 sqrt(1 - "t") "dt"`
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) log(1 + tanx) "d"x`
Evaluate: `int_0^pi 1/(3 + 2sinx + cosx) "d"x`
Evaluate: `int_0^(π/4) sec^4 x dx`
`int_0^(π/2) sin^6x cos^2x.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`
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`
