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∫ tan ^3 (2x)sec (2x) dx give a hint to solve it please shows the beginning steps too

  ∫ tan ^3 (2x)sec (2x) dx  give a hint to solve it 
please shows the beginning steps too

Grade:12

1 Answers

Vikas TU
14149 Points
6 years ago
Dear Student,
For the integrand tan^3(2 x) sec(2 x), substitute u = 2 x and du = 2 dx: =1/2 essential tan^3(u) sec(u) du 
For the integrand tan^3(u) sec(u), utilize the trigonometric personality tan^2(u) = sec^2(u) - 1: 
=1/2 vital tan(u) sec(u) (sec^2(u) - 1) du 
For the integrand tan(u) sec(u) (sec^2(u) - 1), substitute s = sec(u) and ds = tan(u) sec(u) du: =1/2 vital (s^2 - 1) ds 
Incorporating term by term and figuring out constants: 
=1/2 vital s^2 ds - 1/2 vital 1 ds 
vital of s^2 is s^3/3: 
=s^3/6 - 1/2 vital 1 ds 
vital of 1 is s:=s^3/6 - s/2 + consistent 
Substitute back for s = sec(u): 
=(sec^3(u))/6 - (sec(u))/2 + consistent 
Substitute back for u = 2 x: 
=1/6 sec^3(2 x) - 1/2 sec(2 x) + consistent 
Which is equivalent to: 
=> 1/6 sec(2 x) (sec^2(2 x) - 3) + c.
 
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

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