How do you integrate \int e^{\sec x} \sec x\tan x dx ?

[courtrigrad]

Hello all

1. \int 2\sin x + 3\cos x dx Can you even use substitution in this problem. Or can you directly integrate to get \int 2\sin x + 3\cos x dx = -2\cos x + 3\sin x?

2. \int cos^{2} x - cos x dx So u = cos x dx. When we integrate we get \int cos^{2} x - cos x dx = \frac{u^{3}}{3} - \frac{u^{2}}{2} = \frac{(cos x)^3}{3} - \frac{(cos x)^{2}}{2} Is this right?

3. \int sin 2x\ dx. So u = 2x dx Would it be \frac{1}{2} \-cos x?

4. I need help in doing \int e^{\sec x} \sec x\tan x dx

Thanks a lot

 

[dextercioby]

The first is okay,the second is wrong.Try the double angle formula.

Fix the TEX on the last.

Daniel.

 

[arildno]

As for 4., note that \frac{d}{dx}(\frac{1}{\cos{x}})=\frac{1}{\cos(x)}tan(x)

 

[dextercioby]

As for 3.,it should be -\frac{1}{2}\cos 2x + C

And please,remember to put not only the "dx",but also the integration constants...

Daniel.

 

[courtrigrad]

heh sorry for not putting integration constants.

ok so I know that the double angle formulas are: \sin 2x = 2\sin x\cos x and \cos 2x = cos^{2} x - sin^{2} x. I take it that I should use the latter. How would I use this? Couldn't I just do a direct substitution?

Thanks

 

[courtrigrad]

if we have \int x\cos x^{2} dx then u = x^{2} and du = 2xdx. So \int x\cos x^{2} dx = \int \cos x^{2} x dx \frac{1}{2}\int cos x^{2} 2x dx = \frac{1}{2} \ sin x^{2} + C


For \int cos^{2} x - cos x dx I just let u = cos x. Isnt this right?

Thanks

 

[dextercioby]

No.U should use this formula:
\cos^{2}x=\frac{\cos 2x +1}{2}

Daniel.

 

[courtrigrad]

ok i think i got it. is it:

\int cos^{2} x - cos x dx = \frac{\sin 2x}{4} + \frac{x}{2} - \sin x + C?

\cos^2{x} = \frac{1}{2}(\cos 2x + 1)
\int \cos^{2} x dx = \frac{1}{2}\int (\cos 2x + 1) dx
\frac{1}{2}\int \cos 2x dx + \frac{1}{2}\int dx
u = 2x, du = 2dx

Thanks

 

[courtrigrad]

ok for questions like:

\int sec^{2} \frac{x}{2} dx I put u = \frac{1}{2} x, du = \frac{1}{2} dx.. So we have 2\int sec^{2} u du = 2\tan \frac{1}{2}x dx. Is this correct?

For \int tan^{3} x sec^{2} x dx do we use the identity \sec^{2} x = 1 + \tan^{2} x? Is it:

\int tan^{3} x (1 + \tan^{2} x) dx = \int tan^{3} x dx + \int tan^{5} x dx

So \frac{tan^{4} x}{4} + \frac{tan^{6} x}{6} + C

 

[dextercioby]

The last one is not that easy.I can do it using sine and cosine.If u have a better method,then it's only for your good.

Daniel.

 

[dextercioby]

No.What u did was wrong.The integration is wrt "x" and not "tangent of x".

Daniel.

 

[courtrigrad]

well i used the substitution: sec^{2} x = 1 + tan^{2} x

 

[dextercioby]

I could see that.I meant the final 2 integrations.They "stink"...

Daniel.

 

[courtrigrad]

so there is no way to do it with \sec x?

 

[Kelvin]

so there is no way to do it with \sec x?

<br /> \[<br /> \int_{}^{} {e^{\sec x} \sec x\tan xdx} = \int_{}^{} {e^{\sec x} d\left( {\sec x} \right)} = \int_{}^{} {e^\theta d\theta } <br /> \]<br />

 

[dextercioby]

I never said that.Check this out

(\sec x)&#039;=\sec x \tan x (1)

(\tan x)&#039;=\sec^{2}x (2)

I=\int \tan^{3}x \sec^{2}x dx=\int \tan^{2}x \sec x d(\sec x)(3)

Using part integration
I=\tan^{2}x \frac{\sec^{2}x}{2}-\frac{1}{2}\cdot 2 \int \sec^{4}x \tan x dx (4)

For the last integral,use the definitions of the 2 functions and it will be done in a sec.

Daniel.