Variations on Common Inverse Trig-Integrations

The following are variations  of the common Inverse Trigonometric Integrals starting with ArcTan, moving to ArcSec, and concluding with the ArcSin. 

Solving things like the integrals above is relatively "easy"  Once you pick up the pattern.  They usually require only applications of the Pythagorean identity and some creative algebra.  That is…

A² + B² = C² or in trig speak…

Sinθ² + Cosθ² = 1

Which if we divide both sides by Cosθ² can be rearranged as

Tanθ² + 1 = Secθ²

Thus giving us the three identities we need to solve the above problems.

x² + 1 = Tanθ² + 1 = Secθ²

x² – 1 = Secθ² -1 = Tanθ²

1-x² = 1-  Sinθ² = Cosθ²

Thus if we see x² +1 we want to set x = Tanθ with the associated implications that θ = Tan^-1 x and dx = sec²θ.  We do this because if x = Tanθ we can totally rock the identity we derived that Tanθ² + 1 = Secθ².  Similarly if we see things of the form x² – 1 we will want to let x = Secθ with dx = Secθ Tanθ dθ.  And finally, if you see 1-x² let x = Sinθ and let dx = Cosθ.  The use of trig identities was, after all, the entire reason we made the wacky substations in the first place was so we could get some trig identity action working in our favor.  Making these substitutions leaves us with…

(  ___dx___    = (  __ sec²θ  dθ   = (   dθ = θ + C = ArcTan x +C or Tan^-1x + C

 )     x² + 1             )     sec²θ               )

Sweet, letting x = Tanθ was a pretty good idea, especially when combined with that trig identity we tossed in there canceling the Secθ² !  NOTE:   letting X = Tanθ also works if I have 1+X²     because A+B is the same as B+ A.  

This next one though it looks simple is a bit messier.  But let’s dive in.

(  ___dx___   

  )     x² – 1

This time let X = Secθ.  If we do this I can get this to simplify using Secθ² -1 = Tanθ².  If x = Secθ this means θ = Sec^-1x.  It also means that dx = d(Secθ).  I always remember the derivative of secant with the phrase "dseek tan destroy" playing off of the phrase seek and destroy representing the answer to the derivative of secant  = SecθTanθ dθ.  Making these switches leaves us…

(  ___dx___  = (  __ SecθTanθ dθ ___= (  __ SecθTanθ dθ ___=  (  __ Secθ dθ_ =(__1/Cosθ__dθ  

  )     x² – 1        )        Secθ² -1               )              Tanθ²                 )      Tanθ        )  Sinθ/Cosθ  =

 

( __ 1/Sinθ dθ__ = (    CSCθ dθ 

  )                          )                        =Eew Gross!-- use a table,  or magically pull out of your butt the trick of timesing by  __ Cscθ +Cotθ___  

                                        Cscθ + Cotθ 

and using a U substitution where U = the denominator an dU = the numerator. Where the Heck did that come from?  Don’t worry about it, use a table.  Anyway when the dust settles integral solves to

= - ln |csc x + cot x| + C 

Solution Courtesy of http://www.math.com/tables/integrals/more/csc.htm

Note:  It might also be easier to solve Integral with partial fractions.

I can also use the general strategy of letting x = Secθ to solve the more complicated looking integral: 

 (___ dx____

  ) x √(x² -1)

Again let x = Secθ, Sec^-1(x) = θ and dx = SecθTanθdθ.  Making the appropriate substitutions leaves.

(___SecθTanθdθ ____ Using the identity = Secθ² -1 = Tanθ²

  )  Secθ √( Secθ ² -1)

(___SecθTanθdθ _ = (_SecθTanθdθ _ = ∫dθ = θ + C = Sec^-1 (x) +C

 )  Secθ √( Tanθ²)      ) SecθTanθdθ

Finally, let’s explore 

(  ___dx___    This one looks most similar in form to the Sin²θ identity so let x = Sinθ

  )     1-x²

If X = Sinθ then θ = Sin^-1(x) and dx = Cosθ dθ replacing.

(__ Cosθ dθ_  The bottom has an identity I can use to simplify.  1 – Sin²θ = Cos² θ.  Let’s use it.

  )    1-Sinθ²

(__ Cosθ dθ_  =  (__ Cosθ dθ_   = ( __dθ__  = ∫ Secθ dθ = crud!  I have to do more magic tricks.

   )    1-Sinθ²          )     Cosθ²           )   Cosθ     

In this case, the magic trick I need to do is times by ___Secθ +Tanθ___ 

                                                                                Secθ + Tanθ

Where did that come from?  The same bag of fairy dust that let me solve the Cscθ integral we ran into earlier.  Any way, this integral is evaluated by letting U equaling the new bottom U = Secθ + Tanθ and du = Secθ Tanθ + Sec² θ my old integral turns into the new and improved version

∫ du/u = Ln| Secθ + Tan θ | + C Where θ = Sin^-1 (x) which leaves us the nasty looking

Ln| Sec(Sin^-1 (x) ) + Tan (Sin^-1 (x) ) | + C. 

We can draw a trangle and get it a bit more solved and nicer looking, but hey our teacher has a Ph.D in math we’ll let him do the algebra if he wants it.  For our purposes we are done.  But that integral looked horrible so let’s examine a variation of the previous one.

If instead our integral looked like…

  (___ dx____ We could make the same substitution as before let x = Sinθ

   )  √(1-x² )

Thus θ = Sin^_1(x) and dx = Cosθ dθ. The identity 1 – Sinθ² = Cosθ² applies And Voila…

( __ Cosθ dθ__ = ∫dθ =θ + C = Sin^-1 (x) +C

  )    Cosθ

Ah good now that we know

a.     

 (  (___dx___    = ArcTanx + C

)     x² + 1

   (___ dx____ = ArcSec x +C

   ) x √(x² -1)

And (___ dx____ = ArcSin x +C

        )  √(1-x² )

By similar logic we arrived at the variations above.  As a final NOTE: It is useful to remember the identity Cosθ² =(1/2)[ 1+ Cos(2θ)].  This is a great way of drooping a squared exponent to a more manageable linear cos term.