Advertisement
Advertisement
Advertisement
Sum
Evaluate : `int_0^(pi/2) cosx/((1 + sinx)(2 + sin x))*dx`
Advertisement
Solution
Let I = `int_0^(pi/2) cosx/((1 + sinx)(2 + sin x))*dx`
Put sin x = t
∴ cos x·dx = dt
When x = `pi/(2), t = sin pi/(2)` = 1
When x = 0, t = sin 0 = 0
∴ I = `int_0^1 dt/((1 + t)(2 + t)`
= `int_0^1((2 + t) - (1 + t))/((1 + t)(2 + t))*dt`
= `int_0^1[1/(1 + t) - 1/(2 + t)]*dt`
= `int_0^1 1/(1 + t)*dt - int_0^1 1/(2 + t)*dt`
= `[log |1 + t|]_0^1 - [log|2 + t|]_0^1`
= [log(1 + 1) – log(1 + 0)] – [log(2 + 1) – log(2 + 0)]
= log 2 – log 3 + log 2 ...[∵ log 1 = 0]
= `log ((2 xx 2)/3)`
= `log(4/3)`.
Concept: Fundamental Theorem of Integral Calculus
Is there an error in this question or solution?