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Question
Evaluate the following:
`int_(-pi/4)^(pi/4) log|sinx + cosx|"d"x`
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Solution
Let I = `int_(-pi/4)^(pi/4) log|sinx + cosx|"d"x` ......(i)
= `int_(- pi/4)^(pi/4) log|sin(pi/4 - pi/4 - x) + cos(pi/4 - pi/4 - x)|"d"x` ......`[because int_"a" "f"(x) "d"x = int_"a"^"b" "f"("a" + "b" - x) "d"x]`
= `int_(- pi/4)^(pi/4) log|sin(-x) + cosx|"d"x`
= `int_(-pi/4)^(pi/4) log|cosx - sinx|"d"x` ......(ii)
Adding (i) and (ii), we get
2I = `int_(-pi/4)^(pi/4) log|cosx + sinx|"d"x + int_(-pi/4)^(pi/4) log|cosx - sinx|"d"x`
= `int_(-pi/4)^(pi/4) log|(cosx + sinx)(cosx - sinx)|"d"x`
= `int_(-pi/4)^(pi/4) log|cos^2x - sin^2x|"d"x`
∴ 2I = `int_(-pi/4)^(pi/4) log cos2x "d"x`
2I = `2 int_0^(pi/4) log cos 2x "d"x` .....`[because int_(-"a")^"a" "f"(x)"d"x = 2int_0^"a" "f"(x) "d"x "if" "f"(-x) = "f"(x)]`
∴ I = `int_0^(pi/4) log cos 2x "d"x`
Put 2x = t
⇒ dx = `"dt"//2`
Changing the limits we get
When x = 0
∴ t = 0
When x = `pi/4`
∴ t = `pi/2`
I = `1/2 int_0^(pi/2) log cos "t" "dt"` ......(iii)
I = `1/2 int_0^(pi/2) log cos (pi/2 - "t")"dt"`
I = `1/2 int_0^(pi/2) log sin "t" "dt"` ......(iv)
On adding (iii) and (iv), we get,
2I = `1/2 int_0^(pi/2) (log cos "t" + log sin "t")"dt"`
⇒ 2I = `1/2 int_0^(pi/2) log sin "t" cos "t" "dt"`
⇒ 2I = `1/2 int_0^(pi/2) (log 2 sin "t" cos "t")/2 "dt"`
⇒ 2I = `1/2 int_0^(pi/2) (log sin 2"t" - log 2) "dt"`
⇒ 4I = `int_0^(pi/2) log sin 2"t" "dt" - int_0^(pi/2) log 2 "dt"`
Put 2t = u
⇒ 2dt = du
⇒ dt = `"du"/2`
∴ 4I = `1/2 int_0^pi log sin "u" "du" - int_0^(pi/2) log 2 * "dt"`
⇒ 4I = `1/2 xx 2 int_0^(pi/2) log sin "u" "du" - log 2["t"]_0^(pi/2)`
⇒ 4I = `int_0^(pi/2) log sin "u" "du" - log 2 * pi/2`
⇒ 4I = `2"I" - pi/2 log 2` .....[From equation (ii)]
⇒ 2I = `- pi/2 log 2`
⇒ I = `pi/4 log 1/2`
∴ I = `pi/4 log 1/2`.
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