Question

integration of dx/sinx+tanx

Answer 1

Answer:

Step-by-step explanation:

I=∫dxsinx+tanx

Substitute tanx2=t ; dx=2dt1+t2 ; sinx=2t1+t2 ; tanx=2t1−t2

I=∫2dt1+t22t1+t2+2t1−t2=∫1−t22tdt=ln|t|2−t24+C  

=ln∣∣tanx2∣∣2−tan2x24+C

Answer 2

Hope this will help u... :-)