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प्रश्न
Evaluate : `int_0^(pi/2) (1)/(5 + 4 cos x)*dx`
योग
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उत्तर
Let I = `int_0^(pi/2) (1)/(5 + 4 cos x)*dx`
Put `tan(x/2)` = t
∴ x = 2 tan–1 t
∴ dx = `(2)/(1 + t^2)dt`
and
cos x = `(1 - t^2)/(1 + t^2)`
When x = 0, t = 0
When x = `pi/2, t = 1`
∴ I = `((2)/(1 + t^2))/(5 + 4((1 - t^2)/(1 + t^2))dt`
= `int_0^1 (2dt)/(5(1 + t^2) + 4(1 - t^2)dt`
= `int _0^1 2/(5 + 5t^2 + 4 - 4t^2)dt`
= `int_0^1 (2)/(t^2 + 9)*dt`
= `2 int_0^1 1/(t^2 + 3^2)dt`
= `2[1/3 tan^-1 t/3]_0^1`
= `2/3 [tan^(-1) 1/3 - tan^(-1) (0)]`
= `2/3 [tan^(-1) 1/3 - 0]`
= `(2)/(3) tan^-1 (1/3)`.
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