Integral....

**polymerase** · December 23rd 2007, 09:58 AM

Find $\int \frac{dx}{1+4sin^2x}$

thanks!

**galactus** · December 23rd 2007, 10:20 AM

You can make varous subs to tackle this one.

$\int\frac{1}{1+4sin^{2}(x)}dx$

Let $x=tan^{-1}(u), \;\ dx=\frac{1}{u^{2}+1}du$

After you make the subs and do the algebra, you should get:

$\int\frac{1}{5u^{2}+1}du$

Now, make the sub: $u=\frac{tan(w)}{\sqrt{5}}, \;\ du=\frac{\sqrt{5}}{5cos^{2}(w)}dw$

Make the subs and it turns into a very friendly integral:

$\frac{1}{\sqrt{5}}\int{dw}$

Make the resubs and you're done.

**polymerase** · December 23rd 2007, 11:03 AM

Originally Posted by galactus

You can make varous subs to tackle this one.

$\int\frac{1}{1+4sin^{2}(x)}dx$

Let $x=tan^{-1}(u), \;\ dx=\frac{1}{u^{2}+1}du$

After you make the subs and do the algebra, you should get:

$\int\frac{1}{5u^{2}+1}du$

How did u simplify the expression after u made the substitution of $x=tan^{-1}(u)$ ? and how did you "see" this substitution?

**galactus** · December 23rd 2007, 12:36 PM

The simplification is just simple algebra. You shouldn't have any trouble since you're at this level of mathematics. As for 'seeing' the sub, it's a matter of doing them, I supoose.

**Krizalid** · December 23rd 2007, 12:47 PM

Originally Posted by polymerase

Find $\int \frac{dx}{1+4sin^2x}$

$\int {\frac{1} {{1 + 4\sin ^2 x}}\,dx} = \int {\frac{1} {{5\sin ^2 x + \cos ^2 x}}\,dx} = \int {\frac{{\sec ^2 x}} {{5\tan ^2 x + 1}}\,dx} .$

And this is an arctangent, so

$\int {\frac{1} {{1 + 4\sin ^2 x}}\,dx} = \frac{1} {{\sqrt 5 }}\arctan \left( {\sqrt 5 \tan x} \right) + k.$

**polymerase** · December 23rd 2007, 12:55 PM

Originally Posted by galactus

The simplification is just simple algebra. You shouldn't have any trouble since you're at this level of mathematics. As for 'seeing' the sub, it's a matter of doing them, I supoose.

Thanks, i got it...i just fail to see at first that $u=tan\:x$ ! LOL

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