Advertisements
Advertisements
Question
Evaluate: `int_(-4)^2 1/(x^2 + 4x + 13) "d"x`
Advertisements
Solution
Let I = `int_(-4)^2 1/(x^2 + 4x + 13) "d"x`
= `int_(-4)^2 1/(x^2 + 4x + 4 + 9) "d"x`
= `int_(-4)^2 1/((x + 2)^2 + (3)^2) "d"x`
= `[1/3 tan^-1 ((x + 2)/3)]_(-4)^2`
= `1/3[tan^-1(4/3) - tan^-1(-2/3)]`
∴ I = `1/3[tan^-1(4/3) + tan-1(2/3)]` ......[∵ tan−1(− θ) = − tan−1θ]
APPEARS IN
RELATED QUESTIONS
Evaluate: `int_0^1 (x^2 - 2)/(x^2 + 1).dx`
Evaluate the following:
`int_0^a (1)/(x + sqrt(a^2 - x^2)).dx`
`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
`int_0^4 1/sqrt(4x - x^2) "d"x` =
`int_0^(pi/2) log(tanx) "d"x` =
Evaluate: `int_(pi/6)^(pi/3) cosx "d"x`
Evaluate: `int_(- pi/4)^(pi/4) x^3 sin^4x "d"x`
Evaluate: `int_0^1 1/(1 + x^2) "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_0^1(x + 1)^2 "d"x`
Evaluate:
`int_0^(pi/2) cos^3x dx`
Evaluate: `int_0^(pi/4) (tan^3x)/(1 + cos 2x) "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 1/sqrt(3 + 2x - x^2) "d"x`
Evaluate: `int_0^(pi/4) sec^4x "d"x`
Evaluate: `int_0^(pi/2) cos x/((1 + sinx)(2 + sinx)) "d"x`
Evaluate: `int_0^"a" 1/(x + sqrt("a"^2 - x^2)) "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_0^pi x*sinx*cos^2x* "d"x`
Evaluate: `int_0^1 (1/(1 + x^2)) sin^-1 ((2x)/(1 + x^2)) "d"x`
Evaluate: `int_0^(pi/4) (cos2x)/(1 + cos 2x + sin 2x) "d"x`
`int_0^(π/2) sin^6x cos^2x.dx` = ______.
If `int_2^e [1/logx - 1/(logx)^2].dx = a + b/log2`, then ______.
Evaluate:
`int_(-π/2)^(π/2) |sinx|dx`
Evaluate `int_(π/6)^(π/3) cos^2x dx`
Evaluate:
`int_-4^5 |x + 3|dx`
The value of `int_2^(π/2) sin^3x dx` = ______.
Evaluate:
`int_0^(π/2) (sin 2x)/(1 + sin^4x)dx`
`int_0^1 x^2/(1 + x^2)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.
