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How can I integrate dx/1+√sinx from π/6 to π/3? Pls give reply.

How can I integrate dx/1+√sinx from π/6 to π/3? Pls give reply.

Grade:12

1 Answers

Arun
25750 Points
3 years ago
1 / (1 + sin x)
= 1 / ( {sin (x/2) }^2 + {cos (x/2)}^2 + 2 * sin (x/2) * cos x/2)
( considering sin ^2 (x/2) + cos ^ 2 (x/2) = 1 , and, sin 2x = 2 * sin x * cos x )
so, { sin x = 2 * sin (x/2) * cos (x/2) }
so, reducing the above expression,
= 1 / { ( sin (x/2) + cos (x/2) } ^ 2
now divide numerator and denominator by cos^2 (x/2);
= sec ^2 (x/2) dx / { 1 + tan (x/2) } ^2
let us take 1 + tan (x/2) = t…… (I)
dt = 1/2 * sec^2 (x/2) dx
or, 2 * dt = sec^2 (x/2) dx
substituting back ,
= 2 * dt / ( t)^2
= 2 * ( - 1/t) + C
substitute for x, from (I),
= 2 *{ -1 / (1 + tan x/2) } + C1,
= -2 / (1 + tan x/2) + C1….. (ANS)
 

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