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 $$\sec x$$?

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

 

[dextercioby]

I never said that.Check this out

$$(\sec x)'=\sec x \tan x$$ (1)

$$(\tan x)'=\sec^{2}x$$ (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.