# Basic integration

1. Jun 16, 2017

### Tanishq Nandan

1. The problem statement, all variables and given/known data

Integrate: (sinx + cosx)/sqrt(1+sin2x)
2. Relevant equations
Simple trigo formulae:
cos2x=cos^2(x)--sin^2(x)
sin^2(x)+cos^2(x)=1
sin2x=2.sinx.cosx
3. The attempt at a solution
I tried to rationalize the given term,multiplying both numerator and denominator with:
1st time:cosx-sinx
2nd: sqrt(1-sin2x)
3rd: Both the above terms
Everytime,I only got slightly different terms.

So,went to substitution.
But,I'm not finding any suitable term which I can substitute.
If after rationalizing,I substitute sin2x as a variable,say,T,things might have worked out,
EXCEPT that since I multiplied the term both to numerator and denominator,the other term just makes it impossible to express the whole term in terms of a simple integral.
Any hints? (With or without substitution)

2. Jun 16, 2017

### Nidum

Do you know about integration by parts and quotient rule integration by parts ?

Last edited: Jun 16, 2017
3. Jun 16, 2017

### Buffu

$(\cos x + \sin x)^2 = ?$

4. Jun 16, 2017

### Tanishq Nandan

By parts?Yeah,Quotient rule FOR INTEGRATION?I don't think so

5. Jun 16, 2017

### Tanishq Nandan

1+sin2x,I know,but I already told ya
Then,I have a (sinx + cosx) in the denominator as well,right?That's my problem.
I tried that way as well.

6. Jun 16, 2017

### Ray Vickson

If you write 1 as $\sin^2 x + \cos^2 x$, the denominator becomes
$$\sqrt{ \sin^2 x + \cos^2 x + 2 \sin x \cos x} = \sqrt{(\sin x + \cos x)^2}.$$
Can you see how to simplify $\sqrt{(\sin x + \cos x)^2}\:?$

7. Jun 16, 2017

### Tanishq Nandan

Ooo...should have thought of that..
K,got it.Thanks!