Advertisements
Advertisements
Question
Evaluate the following:
`int_0^(pi/2) "dx"/(("a"^2 cos^2x + "b"^2 sin^2 x)^2` (Hint: Divide Numerator and Denominator by cos4x)
Solution
Let I = `int_0^(pi/2) "dx"/(("a"^2 cos^2x + "b"^2 sin^2 x)^2`
Dividing the numerator and denominator by cos4x, we have
I = `int_0^(pi/2) (sec^4x)/(("a"^2 cos^2x)/(cos^2x) + ("b"^2 sin^2x)/cos^2x)^2 "d"x`
= `int_0^(pi/2) (sec^2x * sec^2x)/("a"^2 + "b"^2 tan^2 x)^2 "d"x`
= `int_0^(pi/2) ((1 + tan^2x) sec^2x)/("a"^2 + "b"^2 tan^2 x)^2 "d"x`
Put tan x = t
⇒ sec2x dx = dt
Changing the limits, we get
When x = 0
t = tan 0 = 0
When x = `pi/2`
t = `tan pi/2 = oo`
∴ I = `int_0^oo (1 + "t"^2)/("a"^2 + "b"^2"t"^2)^2 "dt"`
Put t2 = u only for the purpose of partial fraction
∴ `(1 +"u")/("a"^2 + "b"^2"u")^2 = "A"/(("a"^2 + "b"^2"u")) + "B"/("a"^2 + "b"^2"u")^2`
1 + u = A(a2 + b2u) + B
Comparing the coefficients of like terms, we get
a2A + B = 1 and b2A = 1
⇒ A = `1/"b"^2`
Now `"a"^2 * 1/"b"^2 + "B"` = 1
⇒ B = `1 - "a"^2/"b"^2`
= `("b"^2 - "a"^2)/"b"^2`
∴ I = `int_0^oo (1 + "t"^2)/("a"^2 + "b"^2"t"^2)^2`
= `1/"b"^2 int_0^oo "dt"/("a"^2 + "b"^2"t"^2) + ("b"^2 - "a"^2)/"b"^2 int_0^oo "dt"/("a"^2 + "b"^2"t"^2)^2`
= `1/"b"^2 int_0^oo "dt"/("b"^2("a"^2/"b"^2 + "t"^2)) + ("b"^2 - "a"^2)/"b"^2 int_0^oo "dt"/("a"^2 + "b"^2"t"^2)^2`
= `1/"ab"^3 [tan^-1 "t"/("a"/"b")]_0^oo + ("b"^2 - "a"^2)/"b"^2 (pi/4 * 1/("a"^3"b"))`
= `1/"ab"^3 [tan^-1 oo - tan 0] + ("b"^2 - "a"^2)/"b"^2 (pi/(4"a"^3"b"))`
= `1/"ab"^3 * pi/2 + pi/4 * ("b"^2 - "a"^2)/("a"^2"b"^3)`
= `pi/(2"ab"^3) + pi/4 * ("b"^2 - "a"^2)/("a"^3"b"^3)`
= `pi [(2"a"^2 + "b"^2 - "a"^2)/(4"a"^3"b"^3)]`
= `pi/4 (("a"^2 + "b"^2)/("a"^3"b"^3))`
APPEARS IN
RELATED QUESTIONS
If `int_0^alpha(3x^2+2x+1)dx=14` then `alpha=`
(A) 1
(B) 2
(C) –1
(D) –2
By using the properties of the definite integral, evaluate the integral:
`int_0^(pi/2) (2log sin x - log sin 2x)dx`
By using the properties of the definite integral, evaluate the integral:
`int_0^4 |x - 1| dx`
Evaluate: `int_1^4 {|x -1|+|x - 2|+|x - 4|}dx`
Evaluate : `int "x"^2/("x"^4 + 5"x"^2 + 6) "dx"`
`int_0^(pi"/"4)` log(1 + tanθ) dθ = ______
The value of `int_-3^3 ("a"x^5 + "b"x^3 + "c"x + "k")"dx"`, where a, b, c, k are constants, depends only on ______.
f(x) = `{:{(x^3/k; 0 ≤ x ≤ 2), (0; "otherwise"):}` is a p.d.f. of X. The value of k is ______
If f(x) = |x - 2|, then `int_-2^3 f(x) dx` is ______
`int_0^pi sin^2x.cos^2x dx` = ______
The value of `int_2^7 (sqrtx)/(sqrt(9 - x) + sqrtx)dx` is ______
`int_(-pi/4)^(pi/4) 1/(1 - sinx) "d"x` = ______.
`int_(-2)^2 |x cos pix| "d"x` is equal to ______.
`int_0^(2"a") "f"(x) "d"x = 2int_0^"a" "f"(x) "d"x`, if f(2a – x) = ______.
If `int_(-a)^a(|x| + |x - 2|)dx` = 22, (a > 2) and [x] denotes the greatest integer ≤ x, then `int_a^(-a)(x + [x])dx` is equal to ______.
`int_0^1|3x - 1|dx` equals ______.
Let `int ((x^6 - 4)dx)/((x^6 + 2)^(1/4).x^4) = (ℓ(x^6 + 2)^m)/x^n + C`, then `n/(ℓm)` is equal to ______.
Let `int_0^∞ (t^4dt)/(1 + t^2)^6 = (3π)/(64k)` then k is equal to ______.
`int_((-π)/2)^(π/2) log((2 - sinx)/(2 + sinx))` is equal to ______.
`int_-1^1 (17x^5 - x^4 + 29x^3 - 31x + 1)/(x^2 + 1) dx` is equal to ______.
If `int_0^(2π) cos^2 x dx = k int_0^(π/2) cos^2 x dx`, then the value of k is ______.
Evaluate: `int_0^(π/4) log(1 + tanx)dx`.
Evaluate the following limit :
`lim_("x"->3)[sqrt("x"+6)/"x"]`
Evaluate the following definite integral:
`int_1^3 log x dx`
Evaluate the following integral:
`int_-9^9x^3/(4-x^2)dx`
Solve the following.
`int_0^1e^(x^2)x^3dx`
Evaluate the following integral:
`int_0^1x(1-x)^5dx`
Evaluate the following definite integral:
`int_-2^3(1)/(x + 5) dx`
