Question

plz solve karo yar !! ​

Answer 1

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

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Answer 2

Answer:

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\large\sf \blue{Answer:-}$

$\sf \: Let \: I = \int\:sec^4 \: x\: \: dx$

$\sf= \int sec^2 x. sec^2 \: \: x \: \: dx$

$\sf \: = \int ( 1 + tan^2 x ) . sec^2 \: \: x \: \: dx$

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$\sf \pink{putting \: tan \: x \: = t}$

$\sf\implies sec^2 \: xdx \: = dt$

$\sf\therefore I = \int(1 + t^2) dt$

$\sf= \int \: dt + \int t^2 dt$

$\sf = \: t \: + { \frac{t}{3} }^{3} \: + c$

$\sf= tan \: x + \frac{1}{3} \: tan^2 \: x + c \: [ \because \: t \: = \: tan \: x]$

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