Question

integrate

Answer 1

\[\int\limits_{0}^{y} dy/(\sqrt{a^2+y^2)}\]

Answer 2

Are you sure that is the limit of integration?

Answer 3

\[y=atan \theta\]

Answer 4

yes @prealgebra
i am not familiar with the substitution method of integration so will you explain @.Sam

Answer 5

where did that "y "come from???

Answer 6

@Sarkar just to be clear, the problem is\[\int_{0}^{y}{dy\over\sqrt{a^2+y^2}}\]and you do not know trigonometric substitutions at all?

Answer 7

i was doing a sum in physics where i required to integrate the above quantity so thats where i am stumped
the problem is correct cause i double checked my steps in the question in physics???

Answer 8

of you don't know trig subs I don't know how to solve this
so here's an integral table
http://integral-table.com/old/IntegralTable.pdf
I believe this fits form (33)
If you want to run through the actual integral you would use .Sam.'s substitution

Answer 9

\[y=a~\tan \theta\]
\[dy=a ~\sec^2 \theta\]

\[\int\limits_{}^{}\frac{1}{\sqrt{a ^{2}+(a~\tan \theta)^2}}(a~\sec^2 \theta)~d \theta\]
\[\int\limits_{}^{}\frac{(a~\sec^2 \theta)}{\sqrt{a ^{2}(1+\tan^2 \theta)}}~d \theta\]
\[\int\limits_{}^{}\frac{a~\sec^2 \theta}{a\sqrt{(1+\tan^2 \theta)}}~d \theta\]
\[\int\limits_{}^{}\frac{\sec^2 \theta}{\sqrt{(\sec^2~\theta )}}~d \theta\]
\[\int\limits_{}^{} \sec \theta~d \theta\]

Answer 10

\[\int\limits_{}^{} \sec \theta~d \theta\]
\[=\ln(\sec \theta +\tan ~ \theta)\]
\[\theta =\tan^{-1} (\frac{y}{a})\]

\[=\ln(\sec (\tan^{-1} (\frac{y}{a})) +\tan ~ (\tan^{-1} (\frac{y}{a}))\]

Answer 11

okay understood a bit!!thanks a million @.sam

Answer 12

Im not sure my last \[\int\limits_{}^{}\sec \theta ~ d \theta\]
I use calculator to get , ln(secθ+tan θ)
but others is ok

Answer 13

yeah,coincidentally thats the bit which I didnt quite understand!!

Answer 14

@amistre64 can you check my work?

Answer 15

@.Sam. you are correct

Answer 16

ok

Answer 17

but you didn't sub in for sec and tan properly, you need to make a triangle based on your substitution

Answer 18

in case you forgot ho to integrate sec, multiply by (sec+tan)/(sec+tan) and you get a u-sub\[\int\sec\theta d\theta=\int{\sec^2\theta+\sec\theta\tan\theta\over\sec\theta+\tan\theta}d\theta=\ln|\sec\theta+\tan\theta|+C\]now remember the sub...

Answer 19

Yea that's what i've wroted just now ln(secθ+tan θ), and need to substitute θ=tan−1(y/a)

Answer 20

\[y=a\tan\theta\]means this