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Evaluate: ππ∫-π2π2(sin|x|+cos|x|)dx

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प्रश्न

Evaluate: `int_((-π)/2)^(π/2) (sin|x| + cos|x|)dx`

योग
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उत्तर

We have, `int_((-π)/2)^(π/2) (sin|x| + cos|x|)dx`

Let f(x) = sin|x| + cos|x|

Then, f(x) = f(–x)

Since, f(x) is an even function

So, I = `int_((-π)/2)^(π/2) (sin|x| + cos|x|)dx`

= `2int_0^(π/2) (sinx + cosx)dx`

= `2[-cosx + sinx]_0^(π/2)`

= `2[-cos  π/2 + sin  π/2 + cos0 - sin0]`

= 2[0 + 1 + 1 – 0]

= 2(2)

= 4

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2021-2022 (March) Term 2 - Outside Delhi Set 3

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