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प्रश्न
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)
उत्तर
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))`
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