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Question
Integrate the following with respect to x :
`(sin 2x)/("a"^2 + "b"^2 sin^2x)`
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Solution
`int (sin 2x)/("a"^2 + "b"^2 sin^2x) "d"x`
Put a2 + b2 sin2x = u
(0 + b2 × 2 sin x cos x) dx = du
b2 sin 2x dx = du
sin 2x dx = `1/"b"^2 "du"`
`int (sin 2x)/("a"^2 + "b"^2 sin^2x) "d"x = int (1/"b"^2 "du")/"u"`
= `1/"b"^2 log|"u"| + "c"`
`int (sin 2x)/("a"^2 + "b"^2 sin^2x) "d"x = 1/"b"^2 log|"a"^2 + "b"^2 sin^2 x| + "c"`
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