Question

integration cosx by cos3x DX

Answer 1

∫cosx / cos3x.dx 
cos3x= 4 cos³x −3 cos x (Using standard identity) 
=cosx(4 cos²x − 3) 
Hence cosx / cos3x=cosx / cosx(4 cos²x − 3)}=1/(4 cos²x − 3) 
Dividing th enumerator and denominator by cos²x, we get 
1/(4 cos²x − 3)=(1/cos²x)/1/(4 − 3/cos²x ) 
= (sec²x)/(4 − 3sec²x ) Replacing 1/cos²x by sec²x 
=(sec²x)/{4− 3(1 +tan²x )} Replacing sec²x by 1+tan²x in the denominator 
or (sec²x)/{4− 3(1 +tan²x )}= (sec²x)/(1 − 3tan²x ) 
Hence 
∫cosx / cos3x.dx=∫(sec²x)/(1 − 3tan²x )dx ...........(i) 
Let tanx = t →sec²xdx = dt or dx = dt/sec²x 
Substituting in (i), we get 
∫(sec²x)/(1 − 3tan²x )dx=∫(sec²x)/(1 − 3t² )dt/sec²x 
{Resolving 1/(1 − 3t² ) into partial fractions} 
=∫1/(1 − 3t² )dt =1/2 ∫{1/(1 − √3t )+1/(1 + √3t )} dt 
=1/2 {(−1/√3)ln(1 − √3t )+(1/√3)ln(1 + √3t )} +C 
=(1/2√3)ln{(1 + √3t )/(1 − √3t )} +C 
=(1/2√3)ln{(1 + √3tanx )/(1 − √3tanx )} +C 
( Substituting the value of t = tanx)