Question

No one can solve this integral question.
Please find attached photo. Thanks!

Answer 1

No one can solve this question

Answer 2

Try to let x=2secu and hence dx=2tanusecu du

Answer 3

then you get secu/(4*sec^2u+1)du. what next?

Answer 4

after this, transform secu/(4*sec^2u+1) to cosu/(5-sinu)du let a = sinu

Answer 5

Could anyone tell me whether this is a correct ans? a = sqrt(1-4/x^2)

Answer 6

put x=1/t and then solve

Answer 7

from the beginning ? which step? it seems not right

Answer 8

are you familiar with hyperbolic functions?

Answer 9

Only cause I get an inverse hyperbolic tan function in my solution

Answer 10

yes. I see. it seems right.how did you get it? May I know?

Answer 11

Through a similar method as @Yttrium used above. substitution and trig identities. It was a difficult question though I must admit, good challenge though.

Answer 12

after secu/(4*sec^2u+1). what did you do? May I know?

Answer 13

That is what I got from Wolfram Alpha.

Answer 14

1) you can multiply the top and bottom by cos^2(u) to get
INT cos(u)/(cos^2(u)+4)
2) then use basic trig identity to get
INT cos(u)/(5-sin^2(u)
3) substitute s=sin(u)
4)factor out the 5
5)subsitute again (the s^2 / 5)
6)you end up with \[\frac{ 1 }{ \sqrt{5} }\int\limits_{?}^{?}\frac{ 1 }{ 1-(sub letter)^{2} } d(subletter)\]
7) integral of (the bit after the integral sign) is tanh^-1(subletter)
8) sub it all back in

Answer 15

I see this is using hyperbolic of tan. thank you. Actually. I was thinking of integrating
cos(u)/(5-sin^2(u) in terms of log (ln)

Thank you. I have to admit you are very good at math.

Answer 16

You can use logs if you want!! you'll just get a much longer complicated answer :)

Answer 17

yup. indeed.