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Evaluate the following: dxab∫0π2 dx(a2cos2x+b2sin2x)2 (Hint: Divide Numerator and Denominator by cos4x) - Mathematics

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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)

योग
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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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अध्याय 7: Integrals - Exercise [पृष्ठ १६६]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 7 Integrals
Exercise | Q 44 | पृष्ठ १६६

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