Very "Messy" Integrals

**sbhatnagar** · December 18th 2011, 03:05 AM

Math Challenge
Evaluate the following Integrals.

1) $\int \sqrt{x+\sqrt{a^2+x^2}}dx$

2) $\int \sqrt{x+a\sqrt{ax-a^2}}dx$

Moderator approved by CaptainBlack

**princeps** · December 18th 2011, 10:19 AM

Originally Posted by sbhatnagar

Math Challenge
Evaluate the following Integrals.

1) $\int \sqrt{x+\sqrt{a^2+x^2}}dx$

2) $\int \sqrt{x+a\sqrt{ax-a^2}}dx$

1)

$\begin{array}{l}x + \sqrt {{x^2} + {a^2}} = u\\\\\left( {1 + \frac{1}{{2 \cdot \sqrt {{x^2} + {a^2}} }} \cdot 2x} \right)dx = du\\\\\frac{{x + \sqrt {{x^2} + {a^2}} }}{{\sqrt {{x^2} + {a^2}} }}dx = du\\\\dx = \sqrt {{x^2} + {a^2}} \cdot \frac{{du}}{u}\end{array}$

$\begin{array}{l}\sqrt {{x^2} + {a^2}} = u - x\\\\{x^2} + {a^2} = {u^2} - 2ux + {x^2}\\\\2ux = {u^2} - {a^2}\\\\x = \frac{{{u^2} - {a^2}}}{{2u}}\end{array}$

$\begin{array}{l}dx = \sqrt {\frac{{{{\left( {{u^2} - {a^2}} \right)}^2}}}{{4{u^2}}} + {a^2}} \cdot \frac{{du}}{u}\\\\dx = \sqrt {\frac{{{u^4} -2{a^2}{u^2} + {a^4} + 4{a^2}{u^2}}}{{4{u^2}}}} \cdot \frac{{du}}{u}\\\\dx = \frac{{{u^2} + {a^2}}}{{2{u^2}}}du\end{array}$

$\begin{array}{l}\int {\sqrt u \cdot \frac{{{u^2} + {a^2}}}{{2{u^2}}}du = \int {\frac{{{u^2} + {a^2}}}{{2{u^{\frac{3}{2}}}}}du} } \\\\ = \frac{1}{2}\int {{u^{\frac{1}{2}}}} du + \frac{{{a^2}}}{2}\int {{u^{\frac{{ - 3}}{2}}}} du = \\\\\frac{1}{3}{u^{\frac{3}{2}}} - {a^2} \cdot {u^{\frac{{ - 1}}{2}}} + C = \\\\= \frac{1}{3}\sqrt {{{\left( {x + \sqrt {{x^2} + {a^2}} } \right)}^3}} - {a^2}\frac{1}{{\sqrt {x + \sqrt {{x^2} + {a^2}} } }} + C\end{array}$

**Opalg** · December 18th 2011, 12:09 PM

Originally Posted by sbhatnagar

Math Challenge
Evaluate the following Integrals.

1) $\int \sqrt{x+\sqrt{a^2+x^2}}dx$

Substitute $x = a\sinh u$ . Then

$\begin{aligned}\int \sqrt{x+\sqrt{a^2+x^2}}dx &= \int \sqrt{a(\sinh u + \cosh u)}\, a\cosh u\,du \\ &= \tfrac12a^{3/2}\int e^{u/2}(e^u + e^{-u})\,du \\ &= \tfrac12a^{3/2}\int(e^{3u/2} - e^{-u/2})\,du \\ &= \tfrac12a^{3/2}\bigl(\tfrac23e^{3u/2} - 2e^{-u/2}\bigr) + C \\ &= \sqrt{ae^u}\bigl(\tfrac13ae^u - ae^{-u}\bigr) + C \\ &= \sqrt{ae^u}\bigl(\tfrac43a\sinh u - \tfrac23a\cosh u\bigr) + C \\ &= \sqrt{x+\sqrt{a^2+x^2}}\,\bigl(\tfrac43x - \tfrac23\sqrt{a^2+x^2}\bigr) + C. \end{aligned}$

I haven't checked to see whether this answer agrees with princeps's. They may well be equivalent.

**sbhatnagar** · December 19th 2011, 12:05 AM

Your answers are correct are correct!

Hint for Q2)

Spoiler:

**Foxlion** · April 27th 2012, 04:41 PM

in reply to Princeps' signature, it's its own cure!

Thread: Very "Messy" Integrals

LinkBack

Thread Tools

Display

Very "Messy" Integrals

Re: Very "Messy" Integrals

Re: Very "Messy" Integrals

Re: Very "Messy" Integrals

Re: Very "Messy" Integrals

Similar Math Help Forum Discussions

"Mixture" and "Distance, Speed, and Time" Word Problems

What does "solved analytically" and "closed-form solution" means?

Ellipse: extract "minor axis" (b) when given "arc length" and "major axis" (a)

Definition clarification: "state-space", "support" and "sample space"

[SOLVED] Matlab symbolic "solve" gives unknown parameter "z"

Search Tags