# Trig integral

1. May 29, 2008

### Markjdb

1. The problem statement, all variables and given/known data
Find the integral from 0 to pi/2 of sin(x)dx/(cos(x)+sin(x)).

3. The attempt at a solution

You can easily make it into the integral of tan(x)dx/(1+tan(x)), but the substitution u=tanx can't work because tan(pi/2) is undefined. Any ideas for a good substitution?

2. May 30, 2008

### rohanprabhu

Use the identity for definite integrals:

$$\int_a^b f(x)dx = \int_a^b f((a + b) - x)dx$$

and the trignometric identity:

$$\sin\left(\frac{\pi}{2} - x\right) = \cos(x)$$

$$\cos\left(\frac{\pi}{2} - x\right) = \sin(x)$$

Last edited: May 30, 2008
3. May 30, 2008

### Defennder

Where did that identity come from? When does it apply? It doesn't hold for f(x) = x for example.

As for the original integral, it's a little tricky and you have to express it in a special form to arrive at the answer.

$$\frac{sin (x)}{sin (x) + cos (x)} = \frac{1}{2} \ \frac{sin (x) + cos (x) - (cos (x) - sin (x)}{sin (x) + cos (x)} = \frac{1}{2} \left(1 - \frac{cos(x) - sin(x)}{sin(x)+cos(x)} \right)$$

The rightmost expression can be easily integrated.

4. May 30, 2008

### rohanprabhu

Are you kidding me?? The identity is very well valid:

It applies for any function f(x)

really??

$$\int_a^b xdx = \frac{x^2}{2}|_a^b = \frac{b^2}{2} - \frac{a^2}{2}$$

$$\int_a^b (a + b - x)dx = ax|_a^b + bx|_a^b - \frac{x^2}{2}\left|_a^b$$

$$~ =~ a(b - a)~ +~ b(b - a)~ +~ \frac{b^2}{2}~ -~ \frac{a^2}{2}~ = ~ab~ -~ a^2~ + ~b^2~ - ab~ + ~\frac{a^2}{2}~ -~ \frac{b^2}{2}$$

$$= \frac{b^2}{2} - \frac{a^2}{2}$$

even mathematica agrees with me:

5. May 30, 2008

### Defennder

My mistake, I was mentally exhausted when I chanced upon this earlier. But I don't see how this could help evaluate the integral. After applying the transformation, you're left with the same function as before, except this time the top trigo function is cos x.

6. May 30, 2008

### rohanprabhu

That's all right.. happens with all of us :D

The question given above is a VERY common illustration of the above mentioned identity. Let the required integral be 'I'. Then,

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

(eqn. 1)

$$I = \int_0^\frac{\pi}{2} \frac{\sin\left(\frac{\pi}{2} - x\right)}{\sin\left(\frac{\pi}{2} - x\right) + \cos\left(\frac{\pi}{2} - x\right)}dx = \frac{\cos(x)}{\sin(x) + \cos(x)}dx$$

(eqn. 2)

adding both of them together,

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

and hence,

$$I = \frac{\pi}{4}$$

7. May 30, 2008

### DavidWhitbeck

That was very slick Rohan!

8. May 30, 2008

### Defennder

That's something I never did think of. Quite novel indeed.

9. May 30, 2008

### D H

Staff Emeritus
Much nicer than the identity

$$\sin x = \frac 1 2 \bigl((\cos x + \sin x) - (\cos x - \sin x)\bigr)$$

from which

$$\int_0^{\pi/2} \frac{\sin x}{\cos x + \sin x}\,dx = \frac 1 2 \bigl(x - \ln(\cos x + \sin x)\bigr)\Bigl|_0^{\pi/2} = \frac {\pi} 4$$

10. May 30, 2008

### rohanprabhu

which is what Defennder already suggested... and what makes this much nicer??

11. May 30, 2008

### D H

Staff Emeritus
Sorry, my post was too short. Your technique is much nice.