Question

Solve this integration (method of SUBSTITUTION)​

Answer 1

1/6 sec2x (sec2x-3)

put sec2x = t

then 2sec2xtan2x dx = dt

now y = $ [tan^2(2x) × sec2xtan2x dx]

y = (1/2)$ tan^2(2x) dt

y = (1/2)$ (sec^2(2x)-1) dt

y = (1/2)$ (t^2-1) dt

y = (1/2) $(t^2)dt - $dt

y = 1/2 × (t^2)/3 - t

y = 1/6 t ( t-3)

y = 1/6 sec2x (sec2x-3)