How do you integrate int 1/sqrt(-e^(2x)-20e^x-101)dx using trigonometric substitution?

1 Answer

-sqrt(101)/101i*ln ((10((e^x+10)/(sqrt(e^(2x)+20e^x+101)+1))+1-sqrt101)/(10((e^x+10)/(sqrt(e^(2x)+20e^x+101)+1))+1+sqrt101))+C

Explanation:

The solution is a little bit lengthy !!!
From the given int 1/sqrt(-e^(2x)-20e^x-101)*dx

int 1/((sqrt(-1)*sqrt(e^(2x)+20e^x+101))*dx

Take note that i=sqrt(-1) the imaginary number
Set aside that complex number for a while and proceed to the integral

int 1/(sqrt(e^(2x)+20e^x+101))*dx

by completing the square and doing some grouping:
int 1/(sqrt((e^x)^2+20e^x+100-100+101))*dx

int 1/(sqrt(((e^x)^2+20e^x+100)-100+101))*dx

int 1/(sqrt(((e^x+10)^2-100+101)))*dx

int 1/(sqrt(((e^x+10)^2+1)))*dx

First Trigonometric substitution:

The acute angle w with side opposite =e^x+10 and adjacent side =1 with hypotenuse =sqrt ((e^x+10)^2+1)

Let e^x+10=tan w

e^x dx=sec^2 w dw

dx=(sec^2w *dw)/e^x

and then

dx=(sec^2w *dw)/tan (w-10)

The integral becomes

int 1/sqrt(tan^2w+1)*(sec^2w*dw)/(tan w-10)

int 1/sqrt(sec^2w)*(sec^2w*dw)/(tan w-10)

int 1/sec w*(sec^2w*dw)/(tan w-10)

int (secw*dw)/(tan w-10)

from trigonometry sec w=1/cos w and tan w=sin w/cos w
The integral becomes

int (1/cos w*dw)/(sin w/cos w-10) and

int (dw)/(sin w-10 cos w)

Second Trigonometric substitution:

Let w=2 tan^-1 z

dw=2*dz/(1+z^2)

and also z=tan (w/2)

The right triangle: The acute angle w/2 with opposite side = z
Adjacent side =1 and hypotenuse =sqrt (z^2+1)

From Trigonometry : Recalling half-angle formulas

sin (w/2)=sqrt((1-cos w)/2

z/sqrt(z^2+1)=sqrt((1-cos w)/2

solving for cos w

cos w=(1-z^2)/(1+z^2)

Also using the identity sin ^2w=1-cos ^2w
it follows that

sin w=(2z)/(1+z^2)

the integral becomes

int (dw)/(sin w-10 cos w)=int (2*dz/(1+z^2))/((2z)/(1+z^2)-10* (1-z^2)/(1+z^2))

Simplifying the integral results to

int (2*dz)/(2z-10+10z^2)

int ((1/5)*dz)/(z^2+z/5-1)

By completing the square:

int ((1/5)*dz)/(z^2+z/5+1/100-1/100-1)

int ((1/5)*dz)/((z+1/10)^2-101/100)

int ((1/5)*dz)/((z+1/10)^2-(sqrt101/10)^2)

Use now the formula int (du)/(u^2-a^2)=1/(2a)*ln((u-a)/(u+a))+C

Let u=z+1/10 and a=sqrt101/10 and including back the i=sqrt(-1)

Write the final answer using original variables
-sqrt(101)/101i*ln ((10((e^x+10)/(sqrt(e^(2x)+20e^x+101)+1))+1-sqrt101)/(10((e^x+10)/(sqrt(e^(2x)+20e^x+101)+1))+1+sqrt101))+C