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Evaluate: ∫dx2+cosx-sinx

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प्रश्न

Evaluate: `int (dx)/(2 + cos x - sin x)`

योग
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उत्तर

Let I = `int (dx)/(2 + cos x - sin x)`

Put `tan  x/2` = t

⇒ x = 2 tan–1t

∴ dx = `(2 dt)/(1 + t^2)`

And sin x = `(2t)/(1 + t^2)`, cos x = `(1 - t^2)/(1 + t^2)`

∴ I = `int 1/(2 + ((1  -  t^2)/(1  +  t^2)) - ((2t)/(1 + t^2))) * (2dt)/(1  +  t^2)`

= `int (1 + t^2)/(2 + 2t^2 + 1 - t^2 - 2t) * (2dt)/(1 + t^2)`

= `2int 1/(t^2 - 2t + 3) dt`

= `2int 1/((t - 1)^2 + (sqrt(2))^2)  dt`

= `2 xx 1/sqrt(2)  tan^-1  ((t - 1)/sqrt(2)) + C`

= `sqrt(2)tan^-1 ((tan(x/2) - 1)/sqrt(2)) + C`

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2021-2022 (March) Set 1

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