integral sec⁴x dx .solve it
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Rewrite the expression as ∫[(sec ² x) (sec²x)l) ] dx
Recall that sec²x = 1 + tan²x.. Trig ID.
Thereafter, replace 2 in 1, get ∫[(sec ² x) ( 1 + tan²x) ] dx
Multiply the expression to get 1. ∫[(sec²x + sec²x.tan²x) ] dx
Use substitution, say u = tanx .. du = sec²x
and split into two integrals.. ∫ (sec² x + ∫sec². tan ²x) dx Integrate the expression with respect to x, we have ∫sec² x dx + ∫ (u)²du) = tanx + ⅓tan³x + c. Something like that
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