Question #101d6

3 Answers
Jan 11, 2017

#(I)=1/4sec^3xtanx+3/8secxtanx+3/8ln|(secx+tanx)|+C#

Explanation:

#intsec^5xdx#

The trick here is to split the #sec^5x# and use parts

#I=intuv'=uv-intvu'dx#

#color(red)(I=intsec^5xdx)=intsec^3xsec^2xdx#

#u=sec^3x=>du=3sec^3xtanxdx#

#v'=sec^2x=>v=tanx#

#I=sec^3xtanx-int3sec^3xtan^2xdx#

#color(red)(I)=sec^3xtanx-int3sec^3x(sec^2x-1)dx#

#color(red)(I)=sec^3xtanx-int(3sec^5x-3sec^3x)dx#

#color(red)(I)=sec^3xtanx-3color(red)(intsec^5xdx)+3intsec^3xdx#

#color(red)(I)=sec^3xtanx-3color(red)(I)+3intsec^3xdx#

#color(red)(4I)=sec^3xtanx+3intsec^3xdx-------(1)#

#-------------#-

#color(blue)(intsec^3xdx)#

IBP once more

#u=secx=>u'=secxtanx#

#v'=sec^2x=>v=tanx#

#I_2=secxtanx-intsecxtan^2xdx#

#I_2=secxtanx-intsecx(sec^2x-1)dx#

#I_2=secxtanx-(intsec^3x-secx)dx#

#I_2=secxtanx-I_2+intsecxdx#

#2I_2=secxtanx+intsecxdx#

#2I_2=secxtanx+ln|(secx+tanx)|#

#I_2=1/2(secxtanx+ln|(secx+tanx)|)#

now substitute this back into #(1)#
#-------------#-

#4I=sec^3xtanx+3/2(secxtanx+ln|(secx+tanx)|)+C#

#(I)=1/4sec^3xtanx+3/8secxtanx+3/8ln|(secx+tanx)|+C#

Mar 4, 2017

#1/4sec^3xtanx+3/8ln|secx+tanx|+3/8secxtanx+C.#

Explanation:

Let us deal with the Problem in a more 8general way, by finding

#I_n=intsec^nxdx, n in NN.#

Prior to proceeding for #I_n#, let us note that,

#(1) : I_1=intsecxdx=ln|secx+tanx|+c.#

#(2) : I_2=intsec^2xdx=tanx+c.#

For #ngt2, I_n=intsec^nxdx=intsec^(n-2)sec^2xdx, #we use the

IBP :#intuvdx=uintvdx-int{((du)/dx)intvdx}dx.#

#u=sec^(n-2)x rArr (du)/dx={(n-2)sec^(n-3)x}{d/dxsecx}#

#={(n-2)sec^(n-3)}{secxtanx}=(n-2)sec^(n-2)xtanx.#

#v=sec^2x rArr intvdx=tanx.#

#:. I_n=sec^(n-2)xtanx-int[{(n-2)sec^(n-2)xtanx}(tanx)]dx#

#=sec^(n-2)xtanx-(n-2)intsec^(n-2)xtan^2xdx#

#=sec^(n-2)xtanx-(n-2)int{(sec^(n-2)x)(sec^2x-1)}dx#

#=sec^(n-2)xtanx-(n-2)int{sec^nx-sec^(n-2)}dx, i.e.,#

#I_n=sec^(n-2)xtanx-(n-2)I_n+(n-2)I_(n-2)#

#rArr {1+(n-2)}I_n-(n-2)I_(n-2)=sec^(n-2)xtanx, or,#

# (n-1)I_n-(n-2)I_(n-2)=sec^(n-2)xtanx+c, (ngt2, n in NN.)#

Though we have derived the Formula, known as, Reduction

Formula, for #I_n, ngt2,# let us check it for #n=2.#

#:. (2-1)I_2-(2-2)I_0=sec^0xtanx rArr I_2=tanx+c.#

Hence, #I_n# holds good for #n ge 2, n in NN.#

Finally, taking #n=5# in #I_n=I_5=intsec^5xdx,# we have,

#4I_5-3I_3=sec^3xtanx+c_1......(i)#

#n=3 rArr 2I_3-1I_1=secxtanx+c_2#

#rArr 2I_3=ln|secx+tanx|+secxtanx+c_2,...[because,(1)]#

#rArr I_3=1/2ln|secx+tanx|+1/2secxtanx+c_2'.........(ii)#

#:. 4I_5=3(1/2ln|secx+tanx|+1/2secxtanx)+sec^3xtanx#

#:. I_5=1/4[sec^3xtanx+3/2ln|secx+tanx|+3/2secxtanx]#

#I_5=1/4sec^3xtanx+3/8ln|secx+tanx|+3/8secxtanx+C.#

Mar 7, 2017

My apologies, I answered this for the wrong function. Try using a similar method for the given function and see if you can get the right answer.

Explanation:

We can also use the hyperbolic trigonometric functions #sinh(x)# and #cosh(x)# to solve this integral.

Notice the similarity in the following identities:

#{(tan^2(x)+1=sec^2(x)),(sinh^2(u)+1=cosh^2(u)):}#

So if we use the substitution #tan(x)=sinh(u)# then it's also true that #sec(x)=cosh(u)#.

These imply, respectively, that #sec^2(x)dx=cosh(u)du# and #sec(x)tan(x)dx=sinh(u)du#.

Then:

#intsec^3(x)dx=intsec(x)(sec^2(x)dx)=intcosh(u)(cosh(u)du)#

We can use the identity #cosh(2u)=2cosh^2(u)-1#, identical to its non-hyperbolic analog, to say that #cosh^2(u)=1/2(cosh(2u)+1)#.

#=1/2int(cosh(2u)+1)du#

These are both found easily:

#=1/2(1/2sinh(2u)+u)#

The problem then becomes turning these back into expressions of #x#.

First note that:

#1/2sinh(2u)=1/2(2sinh(u)cosh(u))=sinh(u)cosh(u)=tan(x)sec(x)#

Also note that #sinh(x)=(e^x-e^-x)/2# so:

#sinh(u)=(e^u-e^-u)/2=tan(x)#

Solving for #u# yields:

#e^u-e^-u=2tan(x)#

Multiplying through by #e^u# and reordering:

#e^(2u)-e^u(2tan(x))=1#

Completing the square as a polynomial with base #e^u#:

#e^(2u)-e^u(2tan(x))+tan^2(x)=1+tan^2(x)#

Using the trig identity between secant and tangent:

#(e^u-tan(x))^2=sec^2(x)#

#e^u=abs(sec(x)+tan(x))#

#u=lnabs(sec(x)+tan(x))#

Then the original integral is:

#intsec^3(x)dx=1/2(1/2sinh(2u)+u)#

#color(white)(intsec^3(x)dx)=(tan(x)sec(x)+lnabs(sec(x)+tan(x)))/2+C#