Question

Question #101d6

Answer 1

`(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`

Answer 2

`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.`

Answer 3

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`