## CatLove9 3 years ago Integrate. sqrt(1+x^2)/x

1. CatLove9

I think it should be $\int\limits_{}^{} 1/\sin(x)\cos^2x$

2. Mimi_x3

Or start from $\int\frac{\sqrt{1+x^{2}}}{x} dx$

3. CatLove9

You could started from there, I got up to 1/sinxcos^2x

4. Mimi_x3

Assuming that you did the other part correct then.. $\int \frac{1}{sinxcos^2{x}}$ wait..are you sure about that though?

5. lgbasallote

|dw:1342236036378:dw| $\sec \theta = \sqrt{1+ x^2}$ $\tan \theta = x$ $\sec^2 \theta d\theta = dx$ $\int \frac{\sqrt{1+x^2}}{x}dx \implies \frac{\sec \theta(\sec^2 \theta d\theta )}{\tan \theta}$

6. CatLove9

I'm sure that $\int\limits_{}^{} \sec^3x/\tan(x)$

7. CatLove9

That turns into 1/cos^2xsinx

8. lgbasallote

it does?

9. lgbasallote

i mean yeah it does

10. CatLove9

I know I can't do u sub.

11. lgbasallote

there's only one thing you can do...integration by parts

12. lgbasallote

$\int \frac{1}{\sin x \cos^2 x}dx \implies \int \csc x \sec^2 x dx$

13. CatLove9

But can I integrate that?

14. lgbasallote

well i was just spitting out ideas....

15. CatLove9

That's why I didn't change it to that because I knew there was nothing to do there.

16. lgbasallote

well i know i can make 1 = sin^2 x + cos^2 x

17. lgbasallote

$\int \frac{1}{\sin x \cos^2 x}dx \implies \int \frac{ \sin^2 x + \cos^2 x}{\sin x \cos^2 x}dx$

18. CatLove9

OH! That's good

19. lgbasallote

$\int \frac{\sin^2 x}{\sin x \cos ^2 x}dx + \int \frac{\cos^2 x}{\sin x \cos^2 x}dx$

20. lgbasallote

yup saw the answer now :D i'll let you do the good stuff ;)

21. CatLove9

Ah, yes, yes, I see it too.

22. lgbasallote

nice! congrats