# Integrate the following w.r.t.x : 12cosx+3sinx - Mathematics and Statistics

Sum

Integrate the following w.r.t.x : (1)/(2cosx + 3sinx)

#### Solution

Let I = int (1)/(2cosx + 3sinx)*dx

= int (1)/(3sinx + 2cosx)*dx
Dividing numerator and denominator by
sqrt(3^2 + 2^2) = sqrt(13), we get

I = int ((1/sqrt(3)))/(3/sqrt(13) sinx + 2/sqrt(13) cosx)*dx

Since, (3/sqrt(13))^2 + (2/sqrt(13))^2 = (9)/(13) + (4)/(13) = 1,

we take (3)/sqrt(13) =cos oo, (2)/sqrt(13) = sin oo

so that oo = (2)/(3) and oo = tan^-1(2/3)

∴ I = (1)/sqrt(13) int (1)/(sin x + cosoo + cosx sin oo)*dx

= (1)/sqrt(13) int (1)/(sin(x + oo))*dx

= (1)/sqrt(13) int cosec (x + oo)*dx

= (1)/sqrt(13)log|tan|tan((x + oo)/2)| + c

= (1)/sqrt(13)log |tan ((x + tan^-1  2/3)/(2))| + c.

Alternative Method

Let I = int (1)/(2cosx + 3sinx)*dx

Put tan(x/2) = t

∴ x/(2) = tan^-1 t

∴ x = 2tan–1 t

∴ dx = (2)/(1 + t^2)*dt
and
sin x = (2t)/(1 + t^2)
and
cos x = (1 - t^2)/(1 + t^2)

∴ I = int (1)/(2((1 - t^2)/(1 + t^2)) + 3((2t)/(1 + t^2)))*(2dt)/(1 + t^2)

= int (1 + t^2)/(2 - 2t^2 + 6t)*(2dt)/(1 + t^2)

= int (1)/(1 - t^2 + 3t)*dt

= int (1)/(1 - (t^2 - 3t + 9/4) + 9/4)*dt

= int (1)/((sqrt(13)/2)^2 - (t - 3/2)^2)*dt

= (1)/(2 xx sqrt(13)/(2))log |(sqrt(13)/(2) + t - 3/2)/(sqrt(13)/(2) - t + 3/2)| + c

= (1)/sqrt(13)log|(sqrt(13) + 2t - 3)/(sqrt(13) - 2t + 3)| + c

= (1)/sqrt(13)log|(sqrt(13) + 2tan(x/2) - 3)/(sqrt(13) - 2tan(x/2) - 3)| + c.

Is there an error in this question or solution?
Chapter 3: Indefinite Integration - Miscellaneous Exercise 3 [Page 150]

#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 3 Indefinite Integration
Miscellaneous Exercise 3 | Q 3.09 | Page 150

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