Question

Evaluate the following : `int_0^(pi/2) cosx/(3cosx + sinx)*dx`

Answer 1

Let I = `int_0^(pi/2) cosx/(3cosx + sinx)*dx`

Put Numerator = `"A"("Denominator") + "B"[d/dx("Denominator")]`

∴ cos x = `"A"(3cosx + sinx) + "B"[d/dx (3cos x + sinx)]`

= A(3 cos x + sin x) + B(– 3 sin x + cos x)
∴ cos x + 0· sin x = (3A ++ B)cos x (A – 3B) sin x
Comapring the coefficient od sin x and cos x on both the sides, we get
3A + B = 1                                           ...(1)
A – 3B = 0                                           ...(2)
Multiplying equation (1) by 3, we get
9A + 3B = 3                                        ...(3)
Adding (2) and (3), we get
10A = 3

∴ A = `(3)/(10)`

∴ from (1), `3(3/10)  "B" = 1`

∴ B = `1 - (9)/(10) = (1)/(10)`

∴ cos x = `(3)/(10)(3cosx+ sinx) + (1)/(10)(-3sinx + cosx)`

∴ I = `int_0^(pi/2) [(3/10(3cosx + sinx) + 1/10(-3sinx + cosx))/(3cosx + sinx)]*dx`

= `int_0^(pi/2) [3/10 + (1/10 (- 3sinx + cosx))/(3cosx + sinx)]*dx`

= `(3)/(10) int_0^(pi/2) 1*dx + 1/10 int_0^(pi/2) (-3sinx + cosx)/(3cosx  +sinx)*dx`

= `(3)/(10) int_0^(pi/2)+ 1/10 [log|3cosx + sinx|]_0^(pi/2)          ...[because int (f'(x))/f(x)*dx = log int|f(x)| + c]`

= `(3)/(10)[pi/2 - 0] +1/10[log|3 cos  pi/2 + sin  pi/2| - log|3cos 0 + sin0|]`

= `(3pi)/(20) + 1/(10) [log|3 xx 0 + 1| - log|3 xx 1 + 0|]`

= `(3pi)/(20) + 1/10 [log1 - log 3]`

= `(3pi)/(20) - (1)/(10)log3`.             ...[∵ log 1 = 0]