Question

$\displaystyle\bold \red{∫e^{tan(x)} dx}$
Proper solved answer

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Answer 1

Answer:

Let:

I=∫extan(x)dx(1)

Using integration by parts:

It does not matter which of ex and tan(x) you choose to integrate for the first integration by parts, but which you ever you choose you must integrate this term again in any of the following integration by parts or you will end up with the not very useful but true statement I=I.

u=tan(x) du=sec2(x)

v=ex dv=ex

Integration by parts stateu=tan(x) du=sec2(x)

v=ex dv=ex

Integration by parts states that if you have a function u=u(x) and du=u′(x) , while v=v(x) and dv=v′(x) then I is:

I=uv−∫vdu

I=extanx−∫exsec2(x)dx

Note, the following identity:

1+tan2(x)=sec2(x)(2)

I=extanx−∫ex(1+tan2(x))dx

I=extanx−∫exdx−∫extan2(x)dx

I=extanx−ex−∫extan2(x)dx

Now, let us integrate the second part of the integral by calling it I1 and using integration by parts once again:

I1=∫extan2(x)dx

u=tan2(x)du=2tan(x)sec2(x)

v=ex dv=ex

The differential of tan2(x) was obtained using differentiation by chain rule:

u=(tan(x))2

Let:

y=tan(x)

dydx=sec2(x)

u=y2

dudy=2y

dudy⋅dydx=2ysec2(x)

Now, substitute tan(x) back into the equation:

dudx=2tan(x)sec2(x)

Back to I1:

I1=extan2(x)−∫2extan(x)sec2(x)dx

I=extanx−ex−extan2(x)+∫2extan(x)sec2(x)dx

Using Eq. (2):

tan(x)sec2(x)=tan(x)+tan3(x)

I=extanx−ex−extan2(x)+∫2ex(tan(x)+tan3(x))dx

I=extanx−ex−extan2(x)+2∫extan(x)dx+2∫extan3(x)dx

Using Eq. (1):

I=extanx−ex−extan2(x)+2I+2∫extan3(x)dx

Collect like terms:

−II=extanx−ex−extan2(x)+2∫extan3(x)dx=ex−extanx+extan2(x)−2∫extan3(x)dx

Let:

I2=∫extan3(x)dx

Again, I think you know what is coming next, yes you guessed right, integration by parts:

u=tan3(x)du=3tan2(x)sec2(x)

v=ex dv=ex

Differentiation of u is done using the chain rule :

Let:

y=tan(x)

dydx=sec2(x)

u=y3

dudy=3y2

dudy⋅dydx=3y2sec2(x)

Now, substitute tan(x) back into the equation:

dudx=3tan2(x)sec2(x)

Back to I2:

I2=extan3(x)−3∫extan2(x)sec2(x)dx

Using Eq. (2):

tan2(x)sec2(x)=tan2(x)(1+tan2(x))

I2=extan3(x)−3∫extan2(x)(1+tan2(x))dx

I2=extan3(x)−3∫extan2(x)dx−3∫extan4(x)dx

Using I1:

I2=extan3(x)−3(extan2(x)−∫2extan(x)sec2(x)dx)−3∫extan4(x)dx)I2=extan3(x)−3extan2(x)+6∫extan(x)sec2(x)dx

Using Eq. (2):

I2I2I2−5I2I2=extan3(x)−3extan2(x)+6∫ex(tan(x)+tan3(x))dx−3∫extan4(x)dx=extan3(x)−3extan2(x)+6∫extan(x)dx+6∫extan3(x)dx−3∫extan4(x)dx=extan3(x)−3extan2(x)+6I2+6I−3∫extan4(x)dx=extan3(x)−3extan2(x)+6I−3∫extan4(x)dx=−15extan3(x)+35extan2(x)−65I+35∫extan4(x)dx

Therefore,

II−75II=ex−extanx+extan2(x)−2(−15extan3(x)+35extan2(x)−65I+35∫extan4(x)dx)=ex−extanx−65extan2(x)+25extan3(x)−65extan2(x)+125I−65∫extan4(x)dx=ex−extanx−65extan2(x)+25extan3(x)−65∫extan4(x)dx=−57ex+57extanx+67extan2(x)−27extan3(x)+67∫extan4(x)dx

We can see that this integral is not going to end any time soon and cannot be expressed as an elementary function. Integration by parts was a reasonable thing to try but it did not seem to work out, neither will any of the other standard techniques work.

So, let us try something else:

Using Euler’s formula we establish:

eix=cos(x)+isin(x) (3)

e−ix=cos(x)−isin(x) (4)

Therefore,

cos(x)=eix+e−ix/2

isin(x)=eix−e−ix/2

Also, we know that:

tan(x)=sin(x)cosx=−ieix−e−ixeix+e−ix=−i1−e−2ix1+e−2ix=−i−2i∑n=1∞(−1)ne−2inx

I=−iex−2i∑n=1∞(−1)n∫e(1−2ik)xdx=−iex−2i∑n=1∞(−1)n1−2ike(1−2ik)x+Ciex−LerchPhi(−e−2ix,1,1+i2)e(1−2ik)x+C

Answer 2

$\displaystyle\bold \red{∫e^{tanx} \: dx}$

$\displaystyle\bold \red{∫e^{u} \: \frac{1}{ {u}^{2} + 1 } \: du}$

$\displaystyle\bold \red{∫e^{u} \: \: \bigg( \frac{ \frac{1}{2i} }{ui} + \frac{ - \frac{1}{2i} }{u + i} \bigg) \: du }$

$\displaystyle \bold \red{ \frac{1}{2i} \int ei \frac{ {e}^{u - i} }{u - i} - ei \frac{ {e}^{u + i} }{u + i} \: du}$

${\displaystyle \bold \red{ \frac{1.i}{2i.i} \bigg( {e}^{i}Ei(u - i) - {e}^{i} Ei(u + i) \bigg) }}$

$\displaystyle \bf\red{ - \frac{i}{2} ( {e}^{i}Ei( tanx - i) - {e}^{i} Ei(tanx + i)) + C}$