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I want to show that this equation holds for positive a. $$\int \sqrt{a^2 + u^2}\;du = \frac{u}{2} \sqrt{a^2 + u^2}+ \frac{a^2}{2}\;ln(u+\sqrt{a^2+u^2}))+ C $$ My attempt is this: $$u=a.tan(t)\\du=a(1+tan^2(t))dt\\\int \sqrt{a^2 + u^2}\;du=\int \sqrt{a^2 + a^2tan^2(t)}\;du =\int |a.sec(t)|.a.(sec^2(t))\;dt=a^2\int |sec^3(t)|dt$$

And we know that: $$\int sec^3(t)dt=\frac{1}{2}(tan(t).sec(t)+ln|sec(t)+tan(t)|)=I$$ Then I made an attempt to rewrite the right side of my equation. $$\frac{u}{2} \sqrt{a^2 + u^2}= \frac{a.tan(t)}{2}.|a.sec(t)|=\frac{a^2}{2}tan(t)|sec(t)|=II\\\frac{a^2}{2}\;ln(u+\sqrt{a^2+u^2})=\frac{a^2}{2}.ln(a.tan(t)+|a.sec(t)|)=\frac{a^2}{2}ln(a(tan(t)+|sec(t)|))=III$$ If we multiply I by a^2: $$I=\frac{a^2}{2}(tan(t).sec(t))\;[which\;looks\;similar\;to\;II]\\+\frac{a^2}{2}ln|sec(t)+tan(t)|\;[which\;looks\;similar\;to\;III] $$ I'm confused what should I do from now on; My first problem is with absolute values and my second is the fact that III has an "a" element inside the parenthesis of ln but I doesn't. Any help would be appreciated!

Farhad Rouhbakhsh
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1 Answers1

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A natural substitution for this integral is $u=a\sinh t$. Then,

$$\int \sqrt{a^2 + u^2}\;du =a^2\int \sqrt{1+\sinh^2t}\cosh tdt=a^2 \int \cosh^2tdt$$

$$=\frac12 a^2\int (1+\cosh 2t)dt=\frac 12 a^2 (t + \frac12\sinh 2t)+C$$

$$=\frac 12 a^2 \sinh^{-1} \frac ua + \frac 12a^2\sinh t \cosh t+C$$ $$= \frac{a^2}{2}\;\ln(u+\sqrt{a^2+u^2})+ \frac{1}{2}u \sqrt{a^2 + u^2}+C $$

where $\sinh^{-1} x= \ln(x+\sqrt{1+x^2})$ is used.

Edit:

The use of substitution $u=a\tan t$ could be somewhat cumbersome. Continue with

$$I=\int \sec^3t dt= \int (1+\tan^2t)\sec tdt = \int \sec tdt + \int \tan t d(\sec t)$$ $$= \ln(\sec t + \tan t) + \tan t \sec t - I$$

Then,

$$I = \frac12 \ln(\sec t + \tan t) + \frac12 \tan t \sec t + C$$

Then, plug back $\tan t = \frac ua$ and $\sec t = \frac{\sqrt{a^2+u^2}}a$ to recover the given result,

$$a^2\int \sec^3t dt= \frac12a^2 \ln\frac{\sqrt{a^2+u^2}+u}a + \frac12 u\sqrt{a^2+u^2}+ C$$

Quanto
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  • hey man thank you, but help me in my solution. I mean put u=a.tan(t). I want to reach the answer with this. – Farhad Rouhbakhsh Jan 07 '20 at 19:23
  • @FarhadRouhbakhsh - I added your way in the answer as well. – Quanto Jan 07 '20 at 19:40
  • you wrote I the same as what i wrote, but why did you remove absolutes? – Farhad Rouhbakhsh Jan 07 '20 at 19:46
  • @FarhadRouhbakhsh - because $\sec t \ge 1$. – Quanto Jan 07 '20 at 19:47
  • can you explain more. Im newbie in integrals. of course its not true for every t that sec t >=1 – Farhad Rouhbakhsh Jan 07 '20 at 19:49
  • @FarhadRouhbakhsh - The range of secant function is $[1,\infty)$, which is always true. – Quanto Jan 07 '20 at 19:50
  • (sec t = 1 / cos t) and we can put a t such that cos t would be negative. the equation you wrote for secant makes it always positive, but in fact you wrote the equation of $$|sec t|= \sqrt {a^2+u^2} / a $$ by the way what you proved in your second substitution is different from the first. you added a in denominator in the last line but thats not the equation I asked for. – Farhad Rouhbakhsh Jan 07 '20 at 19:55
  • @FarhadRouhbakhsh - I forgot to mention the domain for the substitution is $t\in[-\frac\pi2, \frac\pi2]$. Regarding $a$, note that $-\ln a$ is just a constant, which is to be absorbed into $C$. – Quanto Jan 07 '20 at 20:05
  • I got that absorbing point. but doesnt the fact that t∈[−π/2,π/2] contradict that we are calculating an indefinite integral? and if you have time, please explain me in easy language what is the role of substitution domain in general? – Farhad Rouhbakhsh Jan 07 '20 at 20:10
  • @FarhadRouhbakhsh - Sure. The domain of the original variable is $u\in(-\infty,\infty)$, which, via the substitution $u=a\tan t $, is mapped on to $t\in (-\frac\pi2,\frac\pi2)$. In other words, for every $u$ in $(-\infty,\infty)$, there is a $t$ in $(-\frac\pi2,\frac\pi2)$. The domain of $t$ is confined within $(-\frac\pi2,\frac\pi2)$ – Quanto Jan 07 '20 at 20:17
  • I finally understood and said "aha!". I really appreciate your help. Very very much thanks. Farhad from Iran! – Farhad Rouhbakhsh Jan 07 '20 at 20:21
  • @FarhadRouhbakhsh - You are welcome – Quanto Jan 07 '20 at 20:23