Thread: Integral - Multiplying by a form of one

1. Integral - Multiplying by a form of one

Apparently, I can multiply this by a fraction equal to one that contains some trig functions.

$
\int {\frac{1}
{{1 + \sin x}}dx}
$

I can't seem to find this "trick" multiplier simplify this. I've tried cosx/cosx , cscx/cscx, cotx/cotx, etc.

2. Originally Posted by RedBarchetta
Apparently, I can multiply this by a fraction equal to one that contains some trig functions.

$
\int {\frac{1}
{{1 + \sin x}}dx}
$

I can't seem to find this "trick" multiplier simplify this. I've tried cosx/cosx , cscx/cscx, cotx/cotx, etc.
$\frac{1}{1+\sin x} \cdot \frac{1-\sin x}{1-\sin x} = \frac{1-\sin x}{\cos^2 x} = \sec^2 x - \frac{\sin x}{\cos^2 x}$

3. Originally Posted by ThePerfectHacker
$\frac{1}{1+\sin x} \cdot \frac{1-\sin x}{1-\sin x} = \frac{1-\sin x}{\cos^2 x} = \sec^2 x - \frac{\sin x}{\cos^2 x}$
Thanks. Is there any way to learn how to recognize these patterns better or a certain technique that you use?

4. Originally Posted by RedBarchetta
Thanks. Is there any way to learn how to recognize these patterns better or a certain technique that you use?
A lot of comes from experience.
The more problems you do the more ideas you get to how to solve problems.

5. Here, you have 1+sin(x)

Try to find identities with 1 and sin(x). The first that comes in mind is cos²(x)+sin²(x)=1 ---> 1-sin²(x)=cos²(x)=(1-sin(x))(1+sin(x))

When you're dealing with cot, or tan, it can be interesting to first multiply by cos or sin, regarding the situation.

When you're dealing with a sum of cos or a sum of sin, recall the identities of addition sin(a+b), cos(a+b)... cos(a)+cos(b)=.... sin(a)+sin(b)=....

But it's true, like TPH says, a lot comes with experience.