The integral ∫sec2x/(sec x + tan x)9/2 dx equals (for some arbitrary constant k)

The integral ∫sec2x/(sec x + tan x)9/2 dx equals (for some arbitrary constant k)

Expert Solution

Let I = ∫ sec²x/(sec x + tan x)^(9/2) dx

= ∫sec x .sec x. /(sec x + tan x)^(9/2) dx

Put sec x + tan x = t

sec x tan x + sec²x dx = dt

sec x (tan x + sec x) = dt

sec – tan x = 1/t

2 sec x = t + 1/t

sec x = ½ (t + 1/t)

So I = ½ ∫[(t + 1/t)dt/t]/t^(9/2)

= ½ ∫ [t^(-9/2) + t^(-13/2)] dt

= ½ [ (-2/7)t^(-7/2) – (2/11)t^(-11/2)]+ C

= -[ (t^(-7/2)/7) + t^(-11/2)/11] + C

= -[ (sec x + tan x)^(-7/2)/7 + (sec x + tan x)^(-11/2)/11]+ C

-[ (sec x + tan x)-7/2/7 + (sec x + tan x)-11/2/11]+ C

Nikhil Thakur answered 2 years ago

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Question

The integral ∫sec2x/(sec x + tan x)9/2 dx equals (for some arbitrary constant k)

Solutions

Expert Solution

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