David H. answered • 02/19/17

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I think you use trig substitution

since you have [4x^2+1] that is of form [a^2+1^2] so we let tan(theta) = a = 2x in this problem.

that means sec^2(theta)d(theta) = 2dx

sub back into integral and you have

2pi[1-tan^2(theta)/4]sqrt[tan^2(theta)+1]*sec^2(theta)d(theta/2

simplifying and using fact tan^2y+1 = sec^y

pi{[4-tan^2(theta)]/4}sec(theta)sec^2(theta)d(theta)

=

pi*sec^3(theta)+

(-pi/4)tan^2(theta)sec^3(theta)d(theta)

since you have [4x^2+1] that is of form [a^2+1^2] so we let tan(theta) = a = 2x in this problem.

that means sec^2(theta)d(theta) = 2dx

sub back into integral and you have

2pi[1-tan^2(theta)/4]sqrt[tan^2(theta)+1]*sec^2(theta)d(theta/2

simplifying and using fact tan^2y+1 = sec^y

pi{[4-tan^2(theta)]/4}sec(theta)sec^2(theta)d(theta)

=

pi*sec^3(theta)+

(-pi/4)tan^2(theta)sec^3(theta)d(theta)

Use tan2y = sec2y-1 The second line above becomes

(-pi/4)[-1+sec^2(theta)]sec^3(theta)d(theta)

spit it into two integrals

(5pi/4)sec^3(theta)d(theta) (1*)

(-pi/4)sec^5(theta)d(theta) (2*)

The first integral can be done with integration by parts. Let u = sec(theta) , du = sec(theta)tan(theta)d(theta), and dv = sec^2(theta) so v = tan(theta)

Integral = [sec(theta)tan(theta)]-integral {tan(theta)sec(theta)tan(theta)dtheta}

Use fact tan2y = sec2y-1 so second part becomes

Im letting theta = k now

integral {sec^3(k)dk}= secktank-integral{sec^3k-seck}dk

u can move sec^3k to left and int seck = ln|seck+tank|

2integral {sec^3(k)dk} = sec(k)tan(k)+ln|sec(k)+tan(k)|+C

so taking (5pi/4) into account as well the answer to first integral is

[2/(5pi)][sec(k)tan(k)+ln|sec(k)+tan(k)|] + C ..... (***)

The second can be done with integration by parts as well. Let u = sec(k) so du = sec(k)tank and dv= sec^4(theta) and v = 1+tan^3(k)/3 {u get this by u substitution with u = tank and letting sec^4(k) = (1+tan^2(k))sec^2(k)

so integral = sec(k)[1+tan^3(k)/3]-integral{[1+tan^3(k)/3]sec(k)tan(k)dk}

int tan^4(k)sec(k) =int [sec^2(k)-1]^2sec(k)

= int sec^(5)(k)-2sec^3(k)+sec(k)

integral {sec^5(k)dk} = sec(k)+sec(k)tan^3(k)/3

-integral{sec(k)tan(k)dk}-1/3integral {sec^5(k)-2sec^3(k)+sec(k) dk}

move sec^5k over and the integral of secktank = seck

we also know sec^3k and seck (see integral above)

thus

(4/3)integral{sec^5(k)dk} = sec(k)+sec(k)tan^3(k)/3

-sec(k)+(1/3)[sec(k)tan(k)+ln|sec(k)+tan(k)|]-(1/3)ln|sec(k)+tan(k)]

so simplifying and remembering -pi/4

the second integral is

(-1/pi)[sec(k)tan^3(k)+sec(k)tan(k)] +C.... (*****)

Combine (**** and *****) also sub back in using

tank = 2x

so seck = 1/sqrt(4x^2+1)