How do you solve this integral?

int 3/(4+5sinx) dx

1 Answer
Jun 13, 2018

ln|(2tan(x/2)+1)/(2(tan(x/2)+2))|+C, or,

ln|(2sin(x/2)+cos(x/2))/(2(sin(x/2)+2cos(x/2)))|+C.

Explanation:

Let, I=int3/(4+5sinx)dx.

The substn. to be used for this type is tan(x/2)=t.

:. (sec^2(x/2))(1/2)dx=dt, or, dx=(2dt)/(sec^2(x/2)), i.e.,

dx=(2dt)/(1+tan^2(x/2))=(2dt)/(1+t^2).

Further, 4+5sinx=4+5{(2tan(x/2))/(1+tan^2(x/2))},

=4+5{(2t)/(1+t^2)},

:. 4+5sinx=(4t^2+10t+4)/(1+t^2)={2(2t^2+5t+2)}/(1+t^2).

:. I=3int1/(4+5sinx)dx,

=3int(1+t^2)/{2(2t^2+5t+2)}*2/(1+t^2)dt,

=3int1/(2t^2+5t+2)dt,

=int3/{(t+2)(2t+1)dt,

=int3/{(t+2)2(t+1/2)}dt,

=int(3/2)/{(t+2)(t+1/2)}dt,

=int{(t+2)-(t+1/2)}/{(t+2)(t+1/2)}dt,

=int{(t+2)/{(t+2)(t+1/2)}-(t+1/2)/{(t+2)(t+1/2)}}dt,

=int{1/(t+1/2)-1/(t+2)}dt,

=ln|(t+1/2)|-ln|(t+2)|.

Replacing t" by "tan(x/2), we have,

I=ln|(2tan(x/2)+1)/(2(tan(x/2)+2))|+C,

or, I=ln|(2sin(x/2)+cos(x/2))/(2(sin(x/2)+2cos(x/2)))|+C.

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