How to you integrate (x2-5)1/2
The Attempt at a Solution
I have tried substitution, integration by part, and none seems to work. I really don't know where to start.
Well the cosh substitution is the easiest way, but you can also attack it as follows.How did you get this substitution? Hyperbolic function is not in my school syllabus. Is there any other form of substitution to to solve this question?
Is that meant to be a joke? It obviously leads you around in a big circle back to the original integral.yeah now
\int tan\theta tan\theta sec\theta d\theta
put t = [tex] sec\theta [/tex]
therefore dt = [tex] sec\theta tan\theta d\theta [/tex]
draw a right angled triangle.... using t = [tex] sec\theta[/tex].. you'll get value of two sides... find the value of the third side by pythagoras theorem... that way now you can find [tex] tan\theta [/tex]
put the value of [tex] tan\theta [/tex] and dt in the integrand..... solve !
Hi rahuld.exe. Yes that still has an error, in the triangle you forgot the sqrt sign.sorry i had made a mistake... i forgot the root sign..... but i think this time i got it correct..
could you please check it for me...
PS: moderators... i m not posting the solution... i am actually checking if what i've done is correct...