Homework Help: Integration Questions

1. Feb 8, 2005

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

Last edited: Feb 8, 2005
2. Feb 8, 2005

dextercioby

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

Fix the TEX on the last.

Daniel.

3. Feb 8, 2005

arildno

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

4. Feb 8, 2005

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.

5. Feb 8, 2005

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

6. Feb 8, 2005

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

Last edited: Feb 8, 2005
7. Feb 8, 2005

dextercioby

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

Daniel.

8. Feb 8, 2005

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

Last edited: Feb 8, 2005
9. Feb 8, 2005

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

Last edited: Feb 8, 2005
10. Feb 8, 2005

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.

11. Feb 8, 2005

dextercioby

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

Daniel.

12. Feb 8, 2005

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

13. Feb 8, 2005

dextercioby

I could see that.I meant the final 2 integrations.They "stink"... :tongue2:

Daniel.

14. Feb 8, 2005

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

15. Feb 9, 2005

Kelvin

$$$\int_{}^{} {e^{\sec x} \sec x\tan xdx} = \int_{}^{} {e^{\sec x} d\left( {\sec x} \right)} = \int_{}^{} {e^\theta d\theta }$$$

16. Feb 9, 2005

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.