Question #101d6
3 Answers
Explanation:
The trick here is to split the
IBP once more
now substitute this back into
Explanation:
Let us deal with the Problem in a more 8general way, by finding
Prior to proceeding for
For
IBP :
Though we have derived the Formula, known as, Reduction
Formula, for
Hence,
Finally, taking
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
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
These imply, respectively, that
Then:
#intsec^3(x)dx=intsec(x)(sec^2(x)dx)=intcosh(u)(cosh(u)du)#
We can use the identity
#=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
First note that:
#1/2sinh(2u)=1/2(2sinh(u)cosh(u))=sinh(u)cosh(u)=tan(x)sec(x)#
Also note that
#sinh(u)=(e^u-e^-u)/2=tan(x)#
Solving for
#e^u-e^-u=2tan(x)#
Multiplying through by
#e^(2u)-e^u(2tan(x))=1#
Completing the square as a polynomial with base
#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#