How do you integrate #int (e^x-1)/sqrt(e^(2x) -16)dx# using trigonometric substitution?

2 Answers
Jul 26, 2018

#-i(arcsin(e^x/4)+1/4ln(|(4+sqrt(16-e^(2x)))/e^x|))+c#

Explanation:

#e^x=4sinΘ#
#x=ln(4sinΘ)#
#dx=1/(4sinΘ)*4cosΘdΘ#
#dx=cosΘ/sinΘdΘ#

#int(e^x-1)/(sqrt(e^(2x)-16)##dx#

#int(4sinΘ-1)/(sqrt((4sinΘ)^2-16)##*cosΘ/sinΘdΘ#

#int(4sinΘ-1)/(sqrt(16sin^2Θ-16))##*cosΘ/sinΘdΘ#

#int(4sinΘ-1)/(sqrt(16)*sqrt(sin^2Θ-1)##*cosΘ/sinΘdΘ#

#int(4sinΘ-1)/(4*sqrt(-cos^2Θ)##*cosΘ/sinΘdΘ#

#1/iint(4sinΘ-1)/(4cosΘ##*cosΘ/sinΘdΘ#

#-iint(4sinΘ-1)/(4sinΘ)dΘ#

#-iint((4sinΘ)/(4sinΘ)-(1)/(4sinΘ))dΘ#

#-iint1dΘ+iint1/4cscΘdΘ#

#-iΘ + i/4ln|cscΘ+cotΘ| + c#

#color(red)( bar( ul( | color(white)(a/a) color(black)( -iarcsin(e^x/4)-i/4ln(|(4+sqrt(16-e^(2x)))/e^x|)+c, c in RR ) color(white)(a/a) | )))#

Note 1: We set #e^x# as a form of #sinΘ# because the #e^x# term in the root in the denominator was positive and the constant term was negative.

Note 2: #intcscΘ#=#-ln|cscx+cotx|+c#

Note 3: You get the expression for #Θ# from rewriting the original equation, and you can solve for #cscΘ# and #cotΘ# using a right triangle.

Note 4: There are probably ways to simplify my final answer, but I just kept it like that.

Aug 6, 2018

#I=int(e^x-1)/sqrt(e^(2x)-16)dx#
Let #X=e^x#

#x=ln(X)#

#dx=(dX)/X#

So:

#I=int(X-1)/(Xsqrt(X^2-16))dX#

#=int(cancel(X))/(cancel(X)sqrt(X^2-16))dX-int1/(Xsqrt(X^2-16))dX#

Now let #X=4sec(theta)#

#dX=4sec(theta)tan(theta)d theta#

So :

#I=int(4sec(theta)tan(theta))/sqrt((4sec(theta))^2-16)d theta-int(cancel(4sec(theta))tan(theta))/(cancel(4sec(theta))sqrt((4sec(theta))^2-16))d theta#

#=int(4sec(theta)tan(theta))/sqrt(16(sec(theta)^2-1))d theta-int(tan(theta))/sqrt(16(sec(theta)^2-1))d theta#

Because #sec(theta)^2-1=tan(theta)^2#,

#I=int(cancel(4)sec(theta)cancel(tan(theta)))/(cancel(4tan(theta)))d theta-intcancel(tan(theta))/(4cancel(tan(theta)))d theta#

#=intsec(theta)d theta-1/4int1d theta#

Now, we have #intsec(theta)d theta=ln(|sec(theta)+tan(theta)|)#

Here is a proof if you need the explanation.

So :
#I=ln(|sec(theta)+tan(theta)|)-1/4theta#

Because #theta=sec^(-1)(X/4)=sec^(-1)(e^x/4)#, and #tan(sec^(-1)(u))=sqrt(u^2-1)#, where #u# is a function (here is a proof if you need it.),

#I=ln(|e^x/4+sqrt(e^(2x)/16-1)|)-1/4sec^(-1)(e^x/4)+C#, #C in RR#.

\0/ Here's our answer !