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Evaluate : ∫0π2sinx-cosx1+sinxcosx⋅dx - Mathematics and Statistics

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Sum

Evaluate : `int_0^(pi/2) (sinx - cosx)/(1 + sinx cosx)*dx`

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Solution

Let I = `int_0^(pi/2) (sinx - cosx)/(1 + sinx cosx)*dx`

We use the property, `int_0^a f(x)*dx = int_0^a f(a - x)*dx`.

Here `a = pi/(2)`.

Hence In I, we change x by `pi/(2) - x`.

∴ I = `int_0^(pi/2) (sin(pi/2 - x) - cos(pi/2 - x))/(1 + sin(pi/2 - x) cos(pi/2 - x)`

= `int_0^(pi/2) (cosx - sinx)/(1 + cosx sinx)*dx`

= `- int_0^(pi/2) (sinx - cosx)/(1 + sinx cosx)*dx`

= – I
∴ 2I = 0
∴ I = 0.

Concept: Fundamental Theorem of Integral Calculus
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