# integration problem

• Sep 7th 2009, 11:06 PM
gundu
integration problem
How will you integrate dx
1+sinxcosx
• Sep 7th 2009, 11:17 PM
Defunkt
Quote:

Originally Posted by gundu
How will you integrate dx
1+sinxcosx

Using the fact that: $sin(2 \theta) = 2sin(\theta)cos(\theta)$, we get:

$\int 1+sin(x)cos(x) dx = \int 1 + \frac{1}{2}sin(2x) dx$

Could you finish it off from here?
• Sep 7th 2009, 11:25 PM
Prove It
Quote:

Originally Posted by gundu
How will you integrate dx
1+sinxcosx

Remember that $\sin{2x} = 2\sin{x}\cos{x}$.

Therefore $\sin{x}\cos{x} = \frac{1}{2}\sin{2x}$.

So $\int{1 + \sin{x}\cos{x}\,dx} = \int{1 + \frac{1}{2}\sin{2x}\,dx}$

$= x - \frac{1}{4}\cos{2x} + C$.

Alternatively you could use a $u$ substitution.

$\int{1 + \sin{x}\cos{x}\,dx} = \int{1\,dx} + \int{\sin{x}\cos{x}\,dx}$

$= x + \int{\sin{x}\cos{x}\,dx}$.

Let $u = \sin{x}$ so that $\frac{du}{dx} = \cos{x}$.

Then you get

$= x + \int{u\,\frac{du}{dx}\,du}$

$= x + \int{u\,du}$

$= x + \frac{1}{2}u^2 + C$

$= x + \frac{1}{2}\sin^2{x} + C$

It's not difficult to show that the two answers are the same...
• Sep 7th 2009, 11:51 PM
gundu
thanks

However the question is actually

How will you integrate
1/(1+sinx cosx)

Sorry for the confusion.
• Sep 8th 2009, 12:11 AM
mr fantastic
Quote:

Originally Posted by gundu
thanks

However the question is actually

How will you integrate
1/(1+sinx cosx)

Sorry for the confusion.

Learning something from the previous replies, you should be able to re-write the integral as $\int \frac{2}{2 + \sin (2x)} \, dx$.

Now read this: Wolfram|Alpha

Note that the standard Weierstrass substitution is used: PlanetMath: Weierstrass substitution formulas

You will find several examples if you search these forums using the search tool.

(There are other ways of doing this integral but they rely on clever algebraic parlor tricks).
• Sep 8th 2009, 02:16 AM
red_dog
$\int\frac{1}{1+\sin x\cos x}dx=\int\frac{1}{1+\frac{\sin 2x}{2}}dx=2\int\frac{dx}{2+\sin 2x}$

Now substitute $u=\tan x$.