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प्रश्न
Find `int_ (sin "x" - cos "x" )/sqrt(1 + sin 2"x") d"x", 0 < "x" < π / 2 `
उत्तर
According to question,
let I = `int_ (sin "x" - cos "x" )/sqrt(1 + sin 2"x") d"x", 0 < "x" < π / 2 `
I = `int_ (sin "x" - cos "x")/sqrt(sin^2 "x" + cos^2 "x" + 2 sin "x" .cos "x") d"x"`
= `int_ ( sin "x" - cos "x")/sqrt((sin"x" + cos "x")^2 d"x"`
= `int_ (sin "x" - cos "x")/(sin "x" + cos "x") d"x"`
let sin x + cos x = t
⇒ (cos x - sin x) dx = dt
I = `int_ (-1)/("t") d"t"`
= -ln t + C
= ln `(1/"t") + "C"`
⇒ I = ln `((1)/(sin "x" + cos "x")) + "C"`
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