Solving Integral of sqrt(sinx)/(sqrt(sinx)+sqrt(cosx))dx: Help Needed

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In summary, the Integral ofsqrt(sinx)/(sqrt(sinx) + sqrt(cosx))dx between 0 and pi/2 is equal to pi/4. Using the substitution x=pi/2-y, we get: dy=\frac{2udu}{u^{4}+1}
  • #1
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Before I start, this isn't a homework question.
I'm having trouble solving this question, I was wondering if anyone can help me.
The Question is as follows:

Show that the Integral of sqrt(sinx)/(sqrt(sinx) + sqrt(cosx))dx between 0 and pi/2 is equal to pi/4

Using the substitution x=pi/2-y

From the substitution I have got the (-1) integral of sqrt(cosy)/(sqrt(cosy) +sqrt(siny)) dy between pi/2 and 0 =pi/4

The furthest I have managed to get not including other further failed methods is;

the (1) integral of 1/(1+sqrt(tany))dy between pi/2 and 0

Thanks for any help in advance
Simon
 
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  • #2
And now, let:
[tex]u=\sqrt{\tan(y)}[/tex]
We get:
[tex]\frac{du}{dy}=\frac{1}{2\sqrt{\tan(y)}}\frac{1}{\cos^{2}(y)}=\frac{1}{2u}(\tan^{2}(y)+1)=\frac{u^{4}+1}{2u}[/tex]
Thus, we get:
[tex]dy=\frac{2udu}{u^{4}+1}[/tex]
You may now solve your integral by means of partial fractions decomposition.
 
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  • #3
Well, there's no need for partial fractions. A substitution can work: t=u^2.
 
  • #4
bigubau said:
Well, there's no need for partial fractions. A substitution can work: t=u^2.
Well, 1/(1+u) doesn't look very nice with that substitution.
 
  • #5
Firstly,thanks for the correction on u^4+1/u, we'd noticed you'd missed the 2 out. However we're having problems with solving the partial fractions as we're just not getting a nice answer out on the other side. Also did try the t=u^2 solution but as Arildno said, it's a far more horrible substitution.
 
  • #6
Simon Malzard said:
Firstly,thanks for the correction on u^4+1/u, we'd noticed you'd missed the 2 out. However we're having problems with solving the partial fractions as we're just not getting a nice answer out on the other side. Also did try the t=u^2 solution but as Arildno said, it's a far more horrible substitution.
Note that:
[tex]u^{4}+1=(u^{2}-\sqrt{2}u+1)(u^{2}+\sqrt{2}u+1)[/tex]
 
  • #7
Partial fraction decomposition:
We wish to write:
[tex]\frac{2u}{(1+u)(u^{2}-\sqrt{2}u+1})(u^{2}+\sqrt{2}u+1})}=\frac{A}{1+u}+\frac{Bu+C}{(u^{2}-\sqrt{2}u+1})}+\frac{Du+E}{(u^{2}+\sqrt{2}u+1})}[/tex]
Multiplying with the common denominator, we get:
[tex]2u=A(u^{2}-\sqrt{2}u+1})(u^{2}+\sqrt{2}u+1})+Bu(u^{2}+\sqrt{2}u+1})(1+u)+C(u^{2}+\sqrt{2}u+1})(1+u)+Du(u^{2}-\sqrt{2}u+1})(1+u)+E(u^{2}+\sqrt{2}u+1})(1+u)[/tex]

Equate both sides for each power of "u", to get a system of equations
 
  • #8
No matter wether this is homework or not, it is very like a homework problem. It belongs here.
 
  • #9
As i said earlier, its not a homework question. It's a post exam question and Ugh you're going to love the ingenuity of the lecturer who just showed us a nifty little trick to flush this integration out in a much simpler form in much less time.

The question started as the following:
Show that [tex]\int_0^\frac{\pi}{2}\frac{\sqrt{sin(x)}}{\sqrt{sin(x)} + \sqrt{cos(x)}} = \frac{\pi}{4}dx[/tex]

Lecturers solution:
[tex]I = \int_0^\frac{\pi}{2}\frac{\sqrt{sin(x)}}{\sqrt{sin(x)} + \sqrt{cos(x)}}dx [/tex]
+(by the original substitution)
[tex]J = \int_0^\frac{\pi}{2}\frac{\sqrt{cos(y)}}{\sqrt{cos(y)} + \sqrt{sin(y)}}dy [/tex]

I+J= 2I. From that the solution of both integrals added together between the limits gives [tex]\int_0^\frac{\pi}{2}dx[/tex]

Gives the solution 2I=x between the limits. Which is pi/2. Since this is 2I, divide by 2 to give pi/4. QED
 
  • #10
Very nice! :smile:
 

What is the method for solving this integral?

The method for solving this integral is to use the substitution method, where we substitute u = √(sinx) + √(cosx) and then solve for u.

Can I use any other method to solve this integral?

Yes, you can also use the trigonometric identities to simplify the integral before using the substitution method. This may make the integration process easier.

Why is it necessary to use substitution in this integral?

Substitution is necessary because the integrand contains a square root of trigonometric functions, which cannot be integrated using basic integration rules. By substituting, we can simplify the integral and make it easier to solve.

Is there a specific range of values for the variable x in this integral?

Yes, since we are dealing with trigonometric functions, the range of x is limited to 0 to π/2. Outside of this range, the integral may not converge or may give incorrect results.

What are the steps involved in solving this integral using substitution?

The steps involved are:

  1. Identify a suitable substitution, u = √(sinx) + √(cosx) in this case.
  2. Find the derivative of u with respect to x, du/dx.
  3. Substitute the expression for u and du/dx into the integral.
  4. Simplify the integral using the substitution, and solve for u.
  5. Finally, substitute back the value of u in terms of x to get the final solution.

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