Intergrate: x[(x^(2))-1]^(0.5) dx by subsituting x=(1/cosθ)?

Question

ans: (1/3)[x^(2) -1]^(1.5) + C

Answer 1

x = 1/cos(T)

dx = [-1/cos^2(T)](-sin(T)]dT = [sin(T)/cos^2(T)]dT

x^2 - 1 = [1/cos(T)]^2 - 1 = {[1 - cos^2(T)]/cos^2(T)} = sin^2(T)/cos^2(T)

[x^2 - 1]^(0.5) = sin(T)/cos(T)

Putting all this together gives:

[1/cos(T)][sin(T)/cos(T)][sin(T)/cos^2...

[sin^2(T)/cos^4(T)]dT

Now use;

sin(T)/cos(T) = tan(T) and 1/cos(T) = sec(T)

tan^2(T)sec^2(T)dT

and the fact that d[tan(T)] = sec^2(T)dT

gives the form (w^2)dw

which can be easily integrated to give:

(1/3)tan^3(T) + C

But we want this is terms of x so use:

x = 1/cos(T)

cos(T) = 1/x

T is an angle in a triangle where the adjacent side is 1 and the hypoteneuse is x. So the opposite side is SQRT(x^2 - 1)

tan(T) = SQRT(x^2 - 1)

(1/3)[x^2 - 1]^(3/2) + C

Answer 2

x[(x^(2))-1]^(0.5) = 1/cosθ *[(1/cosθ)^2 - 1]^(0.5)

                       = secθ * [ secθ^2 - 1] ^ 0.5

                       = secθ * [ Tanθ^2 ] ^ 0.5

                       = secθ * tanθ

                      

dx = d ( 1/ cosθ) = d (secθ) = secθ *tanθ dθ

So Integral { secθ * tanθ * secθ *tanθ dθ}

               = integral {secθ^2 *tanθ^2 dθ}

               = integral {secθ^2 *(secθ^2 - 1) dθ}

               = integral {(secθ^4 - secθ^2) dθ}

            

Answer 3

i have no idea ... im guessing the answer is .... 1000

Answer 4

It will be (secTtanT)(sectTtanT)dT= sec^2Tta^2Tdt={(tanT)^3}/3 +c =(1/3) (sec^2T - 1)^(3/2) +c as required

Answer 5

∫ x √(x² - 1) dx =

according to your suggestion, let x = (1/cosθ), that is:

x = secθ →

dx = secθ tanθ dθ

thus, substituting, you get:

∫ x √(x² - 1) dx = ∫ secθ √(sec²θ - 1) secθ tanθ dθ =

recall the trig identity sec²θ = 1 + tan²θ → sec²θ - 1 = tan²θ, thus:

∫ secθ (√tan²θ) secθ tanθ dθ =

∫ secθ tanθ secθ tanθ dθ =

∫ tan²θ sec²θ dθ =

note that this integrand inclused both the function tanθ (even if powered)

and its derivative, that is sec²θ, that is:

∫ tan²θ d(tanθ) =

thus, having both the function tanθ and its differential, you can

integrate it straight into:

[1/(2+1)] tan^(2+1)θ + C =

(1/3) tan³θ + C

now, in order to substitute back, recall:

x = secθ →

tanθ = √(sec²θ - 1) = √(x² - 1)

yielding:

(1/3) tan³θ + C = (1/3)√(x² - 1)³ + C

thus, in conclusion,

∫ x √(x² - 1) dx = (1/3)√(x² - 1)³ + C = (1/3) (x² - 1)^(3/2) + C

I hope this helps....

Bye!

Answer 6

its easy. = 2