Re: How can i evaluate the integral : Pi/2 $(1-3) Sqrt (t^2-
Trig sub is commonly used for this integral or its brethren.
But, I prefer the hyperbolic trig sub with integrals of the form \(\displaystyle \sqrt{t^{2}-a^{2}}\)
\(\displaystyle \int\sqrt{t^{2}-1}dt\)
Let \(\displaystyle t=cosh(u), \;\ dt=sinh(u)du\)
Because \(\displaystyle cosh^{2}(u)-1=sinh^{2}(u)\)
this leads to:
\(\displaystyle \int \sqrt{cosh^{2}(u)-1}sinh(u)du=\int sinh^{2}(u)du\)
\(\displaystyle \frac{1}{2}\int (cosh(2u)-1)du\)
Now, it is easier to finish.
