Question

find integral of sec^4x​

Answer 1

$\Large{\boxed{\underline{\overline{\mathfrak{\star \: AnSwer :- \: \star}}}}}$

$Hey\: there!\\ \\ Solution:\\ \\ \int \:sec^{4}x .dx\\ \\ \int\: sec^{2}x . sec^{2}x .dx\\ \\ Using\: Identity :\\ \\ sec^{2}x\: =\: 1 + tan^{2}x \\ \\ We \:get:\\ \\ \int\: (1 + tan^{2}x ) sec^{2}x .dx\\ \\ Put \:tanx = t....(1) \\ \\ sec^{2}x dx = dt\\ \\ I = \int \: ( 1 + t^{2}) dt\\ \\ = \int\: 1.dt + \int\: t^{2} .dt\\ \\ = t +\dfrac{ t^{3}}{3} + c \\ \\ Using \:equation\: (1) \\ \\ tanx + \dfrac{1}{3} tan^{3}x + C \\ \\ Where \:C\: is\: the \:Arbitrary\: Constant.$

Answer 2

sec^4x = sec^2x(1 + tan^2x)

= sec^2x + sec^2xtan^2x

=> INT = tanx + (tan^3x)/3 + C

write sec^4xtanx as

int sec^3xsecx tanx dx

let secx =. t

secx tanx dx =dt

int t^3dt

= t^4/4 +c

put t = secx

= sec^4x/ 4 +. c