How will you integratedx

1+sinxcosx

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- Sep 7th 2009, 10:06 PMgunduintegration problem
How will you integrate

__dx__

1+sinxcosx - Sep 7th 2009, 10:17 PMDefunkt
- Sep 7th 2009, 10:25 PMProve It
Remember that $\displaystyle \sin{2x} = 2\sin{x}\cos{x}$.

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

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

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

Alternatively you could use a $\displaystyle u$ substitution.

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

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

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

Then you get

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

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

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

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

It's not difficult to show that the two answers are the same... - Sep 7th 2009, 10:51 PMgundu
thanks

However the question is actually

How will you integrate

1/(1+sinx cosx)

Sorry for the confusion. - Sep 7th 2009, 11:11 PMmr fantastic
Learning something from the previous replies, you should be able to re-write the integral as $\displaystyle \int \frac{2}{2 + \sin (2x)} \, dx$.

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, 01:16 AMred_dog
$\displaystyle \int\frac{1}{1+\sin x\cos x}dx=\int\frac{1}{1+\frac{\sin 2x}{2}}dx=2\int\frac{dx}{2+\sin 2x}$

Now substitute $\displaystyle u=\tan x$.