# trig integral 1/(1-sin(x)cos(x))

• Jan 25th 2010, 07:15 PM
oblixps
trig integral 1/(1-sin(x)cos(x))
$\displaystyle \int_{0}^{\pi /2}{\frac{dx}{1-\sin (x)\cos (x)}}$

i saw this integral in a previous post and i'm curious on how to do it. i tried a wierstrauss substitution but that seems to make the problem extremely complicated. i also tried rewriting as $\displaystyle {\frac{dx}{1-{\frac{1}{2}}\sin (2x)}}$ but i'm not sure on what to do next.

what would be the best way to tackle this integral?
• Jan 25th 2010, 07:19 PM
Prove It
Quote:

Originally Posted by oblixps
$\displaystyle \int_{0}^{\pi /2}{\frac{dx}{1-\sin (x)\cos (x)}}$

i saw this integral in a previous post and i'm curious on how to do it. i tried a wierstrauss substitution but that seems to make the problem extremely complicated. i also tried rewriting as $\displaystyle {\frac{dx}{1-{\frac{1}{2}}\sin (2x)}}$ but i'm not sure on what to do next.

what would be the best way to tackle this integral?

Make the substitution

$\displaystyle u = \tan{\frac{x}{2}}$ so $\displaystyle du = \frac{1}{2}\sec^2{\frac{x}{2}}$

and make appropriate transformations for $\displaystyle \sin{x}$ and $\displaystyle \cos{x}$.
• Jan 25th 2010, 09:59 PM
oblixps
yes that is the Weierstrass substitution. i tried that but it seemed to make the problem very complicated. is that the best way to do this problem?
• Jan 25th 2010, 11:01 PM
simplependulum
Quote:

Originally Posted by oblixps
yes that is the Weierstrass substitution. i tried that but it seemed to make the problem very complicated. is that the best way to do this problem?

consider $\displaystyle 1 = \sin^2(x) + \cos^2(x)$

$\displaystyle \int \frac{dx}{ 1- \sin(x)\cos(x) }$

$\displaystyle = \int \frac{dx}{\sin^2(x) - \sin(x)\cos(x) + \cos^2(x) }$

$\displaystyle = \int \frac{ \sec^2(x)~dx }{ \tan^2(x) - \tan(x) + 1}$

Sub. $\displaystyle t = \tan(x) , dt = \sec^2(x)dx$

$\displaystyle I = \int_0^{\infty} \frac{dt}{t^2 - t + 1}$
• Jan 26th 2010, 06:07 AM
wonderboy1953
Multiply numerator and denominator by $\displaystyle 1 + sinxcosx$ and then substitute $\displaystyle cos^2x = 1 - sin^2x$ and you should be able to work your way from there.

$\displaystyle 1 - sin^4x = (1 - sin^2x)(1 + sin^2x)$ and then use partial fractions (just an alternate way of solving).

$\displaystyle 1 + sin^2x = (1 + isinx)(1 - isinx)$ and then use partial fractions again.

[Next time, make all of your ideas into one single post. -K.]
• Jan 26th 2010, 07:29 AM
Pulock2009
[IMG]file:///C:/Users/Guest/Desktop/properties%20of%20definite%20integrals.bmp[/IMG]
i was wondering whether some property of definite integral could be used.