# ∫dx/(sin3x+cos3x)i am using substituition , but the process is getting very lengthy, need a better solution!

Vikas TU
14149 Points
7 years ago
Integerating step by step:
=>  ∫1/sin(3x)+cos(3x)dx
Substitute u=3x ⟶ du/dx=3
=1/3∫1/sin(u)+cos(u)du
∫1/sin(u)+cos(u)du
=∫1/1−tan2(u/2)/tan2(u/2)+1+2tan(u/2tan2(u/2)+1
Substitute v=tan(u/2) ⟶ dv/du=sec2(u/2)/2
=v2+1/2
=−2∫1/v2−2v−1dv
∫1/v2−2v−1dv
Factor of denominator
=∫1/(v−√2−1)(v+√2−1)dv
Apply linearity:
=1/23/2∫1/v−√2−1dv−1/23/2∫1v+√2−1dv
∫1/v+√2−1dv
Substitute w=v+√2−1⟶ dw/dv=1
=∫1/wdw
This is a standard integral:
=ln(w)
Undo substitution w=v+√2−1
=ln(v+√2−1)
Now solving:
∫1v−√2−1dv
Substitute w=v−√2−1⟶ dw/dv=1
=∫1/wdw
Use previous result:
=ln(w)
Undo substitution w=v−√2−1
=ln(v−√2−1)
Plug in solved integrals:
1/23/2∫1/v−√2−1dv−1/23/2∫1/v+√2−1dv
=ln(v−√2−1)232−ln(v+√2−1)232
Plug in solved integrals:
−2/∫1v2−2v−1dv
=ln(v+√2−1)√2−ln(v−√2−1)√2
Undo substitution v=tan(u/2)
=ln(tan(u/2)+√2−1)√2−ln(tan(u/2)−√2−1)√2
Plug in solved integrals:
1/3∫1/sin(u)+cos(u)du
=ln(tan(u/2)+√2−1)−ln(tan(u/2)−√2−1)/3⋅√2
Undo substitution u=3xu=3x:
=ln(tan(3x/2)+√2−1)−ln(tan(3x2)−√2−1)/3⋅√2