Question

Evaluate the following:

`int_(-pi/4)^(pi/4) log|sinx + cosx|"d"x`

Answer 1

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`.