हिंदी

Evaluate: ππ∫-π/4π/4cos2x1+cos2xdx.

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प्रश्न

Evaluate: `int_(-π//4)^(π//4) (cos 2x)/(1 + cos 2x)dx`.

योग
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उत्तर

`int_(-π//4)^(π//4) (cos 2x)/(1 + cos 2x)dx = int_(-π//4)^(π//4) (2 cos^2 x - 1)/(2 cos^2 x)dx`

= `1/2 . 2 int_0^(π//4) (2 - sec^2 x)dx`  ...[even function]

= `1/2 . 2[2x - tan x]_0^(π//4)`

= `π/2 - 1`

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2022-2023 (March) Delhi Set 2

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