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A Trigonometry Integral Requiring Two Substitutions

Date: 05/03/98 at 17:14:16
From: Paul.
Subject: Calculus Integration.

What is the integral of

   sqrt(1 + sin(x)), 

where sqrt stands for "square root of"?


Date: 05/04/98 at 16:58:23
From: Doctor Sam
Subject: Re: Calculus Integration.


This is a tricky problem. It will take (I think) two different 
substitutions. We want to find:

   INT sqrt(1 + sin x) dx

I am going to try substituting u = sin(x) to try to remove the trig 
function. When you make a substitution, you must also substitute for 
dx. So:

   u = sin(x)     and     du = cos(x) dx

This gives dx = du/cos(x), and changes the integral to  

       sqrt(1 + u) 
   INT ----------- du

This is no good. We need to get an integral in terms of the u variable 
alone. Here's where a little right-triangle trigonometry can help. We 
made the substitution u = sin(x), so we can visualize a triangle with 
an acute angle x whose sine is u.

Here is one such triangle:

        / |
       /  |
      /   |
   1 /    |  u
    /     |
   /x     |

Now we can use the Pythagorean Theorem to find the third side, and 
then the cosine of x. The third side is sqrt(1 - u^2), and so:

   cos(x) = sqrt(1 - u^2)

Our integral is now:

        sqrt(1 + u) 
   INT ------------- du
       sqrt(1 - u^2)

Now I can't help but notice that

   1 - u^2 = (1 - u)(1 + u)

so this fraction simplifies to:

   INT ----------- du
       sqrt(1 - u)

We are almost done. We have now transformed our trig integral into an 
algebraic integral. Now a second substitution:

   w = 1 - u 

should finish the job.

If w = 1 - u, then dw = -du, so du = -dw. This gives:

   - INT ------- dw

Interpret this as w^(-1/2), and we can use the formula for 
antidifferentiating u^n:

   - INT ------ dw  = -2w^(1/2) + C

Now change back from w to u using w = 1 - u:

   -2w^(1/2) = -2 sqrt(1 - u) + C

And now change back from u to x using u = sin(x):

   -2w^(1/2) = -2 sqrt(1 - u) + C = -2 sqrt(1 - sin(x)) + C

I hope that helps.

Doctor Sam, The Math Forum   

Date: 05/08/98 at 00:26:54
From: Paul Oommen
Subject: Re: Calculus Integration.

Thanks for the answer.

Associated Topics:
High School Calculus

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