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

[Simon43254]

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

 

[arildno]

And now, let:
u=\sqrt{\tan(y)}
We get:
\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}
Thus, we get:
dy=\frac{2udu}{u^{4}+1}
You may now solve your integral by means of partial fractions decomposition.

 

[dextercioby]

Well, there's no need for partial fractions. A substitution can work: t=u^2.

 

[arildno]

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.

 

[Simon43254]

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.

 

[arildno]

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:
u^{4}+1=(u^{2}-\sqrt{2}u+1)(u^{2}+\sqrt{2}u+1)

 

[arildno]

Partial fraction decomposition:
We wish to write:
\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})}
Multiplying with the common denominator, we get:
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)

Equate both sides for each power of "u", to get a system of equations

 

[Integral]

No matter wether this is homework or not, it is very like a homework problem. It belongs here.

 

[Simon43254]

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 \int_0^\frac{\pi}{2}\frac{\sqrt{sin(x)}}{\sqrt{sin(x)} + \sqrt{cos(x)}} = \frac{\pi}{4}dx

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

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

Gives the solution 2I=x between the limits. Which is pi/2. Since this is 2I, divide by 2 to give pi/4. QED

 

[arildno]

Very nice!