indefinite integral

**U-God** · November 2nd 2008, 03:59 PM

I seem to be getting worse and worse at integrating haha.. I can get this in a form that looks simple but for the life of me cannot seem to get past it!!

$\int \frac{sin(kx)}{\sqrt{2+cos(kx)}}dx$ where k does not equal zero

let $cos(kx) = u$ therefore $dx = \frac{du}{-ksin(kx)}$

so it becomes $-\frac{1}{k} \int \frac{du}{\sqrt{2 + u}}$

for some reason I cannot get past here.. Very frustrating!!
Thanks in advance,

**Chris L T521** · November 2nd 2008, 04:01 PM

Originally Posted by U-God

I seem to be getting worse and worse at integrating haha.. I can get this in a form that looks simple but for the life of me cannot seem to get past it!!

$\int \frac{sin(kx)}{\sqrt{2+cos(kx)}}dx$ where k does not equal zero

let $cos(kx) = u$ therefore $dx = \frac{du}{-ksin(kx)}$

so it becomes $-\frac{1}{k} \int \frac{du}{\sqrt{2 + u}}$

for some reason I cannot get past here.. Very frustrating!!
Thanks in advance,

Try this substitution [for the original integral]:

Let $u=\cos(kx){\color{red}+2}$

See how you go with this one

--Chris

**U-God** · November 2nd 2008, 04:03 PM

Originally Posted by Chris L T521

Try this substitution [for the original integral]:

Let $u=\cos(kx){\color{red}+2}$

See how you go with this one

--Chris

you're a genius
thanks tons

**U-God** · November 2nd 2008, 04:12 PM

Actually one more while we're at it would be fantastic!

$\int \frac{1}{5x - \sqrt{x}}dx$
I don't really know how to begin this one, I tried looking for a substitution but couldn't see one that work (although I may be blind).

Cheers,

**Chris L T521** · November 2nd 2008, 04:26 PM

Originally Posted by U-God

Actually one more while we're at it would be fantastic!

$\int \frac{1}{5x - \sqrt{x}}dx$
I don't really know how to begin this one, I tried looking for a substitution but couldn't see one that work (although I may be blind).

Cheers,

Note that $\frac{1}{5x-\sqrt{x}}=\frac{1}{\sqrt{x}\left(5\sqrt{x}-1\right)}$

Then make the substitution $u=5\sqrt{x}-1$

See how you go with this one

--Chris

**U-God** · November 2nd 2008, 04:27 PM

I can never see things like you do!! Cheers!

**U-God** · November 2nd 2008, 05:02 PM

Okay, I got one here that I thought I'd done right but it doesn't match the answer:

$\int \sqrt{1 + 4x^2} dx$ so I let $x = \frac{1}{2}sinh(u)$
therefore $dx = \frac{1}{2}cosh(u)du$

subbed it in: $\frac{1}{2}\int cosh^2(u)du$

used identity to re-arrange: $\frac{1}{4} \int cosh(2u)du + \frac{1}{4}\int du$

evaluated: $\frac{1}{8}sinh(u) + \frac{1}{4}u + C$

re arranged the substitution I made earlier for $u = arcsinh(2x)$
subbed u back in yielded $\frac{1}{8}sinh(arcsinh(2x)) + \frac{1}{4}arcsinh(2x) + C$

which cancelled to give $\frac{1}{4}x + \frac{1}{4} arcsinh(2x) + C$

however the answer I'm given is: $\frac{1}{4}(arcsinh(2x) + 2x\sqrt{1 + 4x^2}) + C$

Hopefully someone can spot what I've done wrong. Thanks!

**mr fantastic** · November 2nd 2008, 05:13 PM

Originally Posted by U-God

Okay, I got one here that I thought I'd done right but it doesn't match the answer:

$\int \sqrt{1 + 4x^2} dx$ so I let $x = \frac{1}{2}sinh(u)$
therefore $dx = \frac{1}{2}cosh(u)du$

subbed it in: $\frac{1}{2}\int cosh^2(u)du$

used identity to re-arrange: $\frac{1}{4} \int cosh(2u)du + \frac{1}{4}\int du$

evaluated: $\frac{1}{8}sinh(<font color=$ 2u) + \frac{1}{4}u + C " alt=" \frac{1}{8}sinh(2u) + \frac{1}{4}u + C " /> Mr F added the red 2.

re arranged the substitution I made earlier for $u = arcsinh(2x)$
subbed u back in yielded $\frac{1}{8}sinh(arcsinh(2x)) + \frac{1}{4}arcsinh(2x) + C$

which cancelled to give $\frac{1}{4}x + \frac{1}{4} arcsinh(2x) + C$

however the answer I'm given is: $\frac{1}{4}(arcsinh(2x) + 2x\sqrt{1 + 4x^2}) + C$

Hopefully someone can spot what I've done wrong. Thanks!

You should ask new question in a new thread.

I think you've made the same mistake (missing the 2) as you did here: http://www.mathhelpforum.com/math-he...integrals.html

Use the same approach.

**U-God** · November 2nd 2008, 05:22 PM

Originally Posted by mr fantastic

You should ask new question in a new thread.

[...]

Will do, was trying not to flood the forum with my threads though

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