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प्रश्न
Evaluate: `int_0^pi e^cosx/(e^cos x + e^-cos x) dx`
मूल्यांकन
उत्तर
Let I = `int_0^pi e^cos x/(e^cos x + e^-cos x) dx` ...(i)
∴ I = `int_0^pi e^cos(pi-x)/(e^cos(pi-x) + e^-cos(pi - x)) dx ...["Using" int_0^a f(x) dx = int_0^a f(a - x)dx]`
or, I = `int_0^pi e^-cos x/(e^-cos x + e^ cos x)` dx ...(ii)
On adding eqs. (i) & (ii), we get
2I = `int_0^pi (e^cos x + e^-cos x)/(e^cos x + e^-cos x) dx`
= `int_0^pi dx`
= `[x]_0^pi = pi`
∴ I = `(pi)/2`
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