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Sum
Find: `int_ (cos"x")/((1 + sin "x") (2+ sin"x")) "dx"`
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Solution
`int_ (cos"x")/((1 + sin "x") (2+ sin"x")) "dx"`
Put 2 + sin x = t
⇒ 1 + sin x = t - 1
cos x dx = dt
`int_ ("dt")/(("t" -1) "t")`
= `int_ ((1)/("t" - 1) - (1)/("t")) "dt"`
= `int_ (1)/("t" -1) "dt" - int 1/"t" "dt"`
= log (t - 1) - log t + C
= log (2 + sin x - 1) - log (2 + sin x) + C
= log (1 + sin x) - log (2 + sin x) + C
= `"log" ((1+ sin "x")/(2 + sin "x")) + "C" ` ...`(∵ "log m" - "log n" = "log" ("m"/"n"))`
Concept: Integration Using Trigonometric Identities
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