Integration by Trignometric substitutions

Introduction to trigonometric substitutions :  Trignometric substitutions enable us to replace the binomials a^2 +x^2 , a^2-x^2 , and x^2-a^2  by single squared terms and thereby transform a number of integrals containing square root s into integrals we can evaluate directly .
The most common substitutions are x = a  tan (theta ) ,
x = a sin (theta) and
x = a sec (theta).
With  x = a tan (theta) ,
a^2 + x^2 = a^2 + a^2 tan ^2(theta)
= a^2 (1 + tan ^2 (theta))
= a^2 vsec^2 (theta).

With a = a sin (theta) ,
a^2-x^2 = a^2 – a^2 sin^2(theta)
= a^2 (1-sin^2(theta))
= a^2cos^2(theta).

With x =a sec(theta) ,
x^2 – a^2 = a^2 sec^2(theta) – a^2
= a^2(sec^2(theta) -1)
= a^2tan^2(theta).

Integration by Trigonometric Substitution :  we want any substitution we use  in an integration to be reversible so that we can change back to the original variable afterward .
for example if x = a tan (theta) ,
we want be able to set (theta) = tan ^-1(x/a) after the integration takes place .
If x = a sin (theta) , we want be able to  set (theta) = sin^-1 (x/a) ,
where we are done similarly for x = a sec (theta)To simplify calculations with the substitution x = a sec(theta) , we will restrict its use to integrals in which (x/a)=1.

This will place (theta) in and make tan (theta)= 0 .
We will  then have
sqrt (x^2 –a^2) = sqrt (a^2 tan^2(theta))
= mod(a tan (theta))= a tan (theta),
free of absolute values , provide a >0.

Trigonometric Substitution Examples :Let us take a  Trigonometric Substitution Problems to understand Integration by Trigonometric Substitution  Here we have to evaluate  integration (dx / sqrt (4 +x^2)).

we can see from the  problem we have to use Trigonometric Substitution Integration .
let us  set x = 2 tan (theta) ,
dx = 2 sec^2 (theta) d(theta), -pi/2< theta4 +x^2 = 4+4 tan^2 (theta)
= 4 (1+ tan^2 (theta) )= 4 (sec^2 (theta)).

Then  integration dx / sqrt (4+x^2) = integration 2sec^2
(theta)d(theta) / sqrt 4sec^2 (theta) = integration sec^2(theta)d(theta)
=  mod (sec (theta)

and we know sqrt sec^2 (theta) = mod (sec(theta) , sec (theta)> 0  for –pi/2 < theta < pi/2 .
In mod (sec (theta)+ tan (theta) +c
= In mod (sqrt (4 +x^2 )/2 + x/2) + c
= In mod (sqrt (4 +x^2) +x )+C’