# value of ∫dx/√3  sinx+cosx is equal to=? give the process of it also.plz reply soon!

sethurathienam Iyer
16 Points
12 years ago

Multiply and divide by 2.

I= dx/root(3) sinx+cosx

then 2I= dx/(root(3)/2 *sinx + (cosx)/2 )

then root(3)/2=cos30 1/2 = sin 30

then root(3)sinx+ cosx becomes  2sin(30+x) .  then the integration becomes integration of cosec(x+30) multiplied by 1/2.

which equals  [log(tan(x+30)/2))] . 1/2+constant.

HOPE I HAVE NOT CONFUSED U!

saket shrivastava
36 Points
12 years ago

(√3)i=∫(sinx+cosx)dx

(√3)i=(-cosx+sinx) + c

no complexity now convet into anythin u wantt

Tanya
11 Points
6 years ago
I=∫dx/(cosx + root3*sinx )
so , 2I=∫dx/ (cosx/2 +root3/2 sinx)             just multiply and divide by 2
2I=∫dx/(cosx * cos 60  + sin60 *sinx)          because cos60=1/2 and sin60=root3/2
2I=∫dx/cos(x-60)                                        because cosa*cosb+sina*sinb= cos(a-b)
2I=∫sec(x-60) dx                                        because 1/cosa = seca
I=1/2∫sec(x-60) dx
I= ½ Log[ | tan(x-60) + sex (x-60 ) | ] + C

3 years ago
Dear student,

I = ∫dx/(cosx + 3 sinx )
so , 2I = ∫dx/ (cosx/2 + 3/2 sinx)             {multiply and divide by 2}
2I = ∫dx/(cosx * cos 60  + sin60 *sinx)
2I = ∫dx/cos(x – 60)                                        {Since, cosa*cosb + sina*sinb = cos(a – b)}
2I = ∫sec(x – 60) dx                                        { 1/cosa = seca }
I = 1/2∫sec(x – 60) dx
I =  ½ log[ | tan(x – 60) + sex (x – 60 ) | ] + C

Thanks and regards,
Kushagra