Question

integration x power 20 sec power 2 dx​

Answer 1

Answer:

∫sec4xdx

=∫sec2x(sec2x)dx

=sec2xtanx−∫tanx(2secx)(secxtanx)dx

=sec2xtanx−2∫tan2xsec2xdx

=sec2xtanx−2∫(sec2x−1)sec2xdx

=sec2xtanx−2∫(sec4x−sec2x)dx

=sec2xtanx−2sec4xdx+2∫sec2xdx

⟹3∫sec4xdx=sec2xtanx+2tanx

⟹∫sec4xdx=13sec2xtanx+23tanx

=13sec2xtanx+23tanx+C

Mulenga Mumba Jr.

Answered November 6, 2018

To integrate ∫sec⁴x dx…

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.