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Evaluate : ∫0π4cosx4-sin2x⋅dx - Mathematics and Statistics

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Question

Evaluate : `int_0^(pi/4) (cosx)/(4 - sin^2x)*dx`

Sum

Solution

Let I = `int_0^(pi/4) (cosx)/(4 - sin^2x)*dx`

Put sin x = t
∴ cos x·dx = dt

When x = `pi/(4), t = sin  pi/(4) = (1)/sqrt(2)`

When x = 0, t = sin 0 = 0.

∴ I = `int_0^(1/sqrt(2)) dt/(2^2 - t^2)`

= `[1/(2(2)) log|(2 + t)/(2 - t)|]_0^((1)/sqrt(2))`

= `(1)/(4)[log((2 + 1/sqrt(2))/(2 - 1/sqrt(2))) - log((2 + 0)/(2  - 0))]`

= `(1)/(4)[log((2sqrt(2) + 1)/(2sqrt(2) - 1)) - log 1]`

= `(1)/(4)log((2sqrt(2) + 1)/(2sqrt(2) - 1))`.       ...[∵ log 1 = 0]

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Fundamental Theorem of Integral Calculus
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Chapter 4: Definite Integration - Exercise 4.2 [Page 172]

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