find the indefinite integral

Question

find the indefinite integral:

∫dx/sinx+tanx

Answer 1

If we let t = tan ( x/2 ) , - π < x < π , then by sketching a right triangle we get

cos( x/2 ) = 1/ √(1+t²) and sin ( x/2 ) = t/ √(1+t²).

Then by using the double angle formulas we obtain

cos( x ) = (1 - t² )/ (1+ t² ) , sin ( x ) = 2t / (1+ t² ) and tan( x ) = 2t / (1- t² )

Also d x = [ 2/ ( 1+ t² )] d t.

Substituting the above to the original integral we can convert into an ordinary rational function of t.

Therefore ∫ dx / ( sin x + tan x ) = ∫ [2 / ( 1+ t² ) d t ] / [ 2 t / ( 1+ t² ) + 2 t / ( 1- t² ) ] =

=( 1/2 ) ∫[( 1- t² ) d t ] / t = (1 /2 ) ∫(d t)/t - (1/2) ∫t d t = ( 1/ 2) ln t - ( 1/4 ) t² +ξ =

= ( 1/ 2) ln [tan(x/2)] - ( 1/4 ) [ tan(x/2)]² +ξ , where ξ is a constant.

Thanks for the question. This was a "juicy" one.

Answer 2

If you mean ∫dx/(sin x + tan x),

I would suggest you use the formulas to convert this into a rational fraction which start with u=tan(x/2).

Using that method I get

ln tan(x/2)-(1/2)tan²(x/2) + C