Question

Calculus Trigonometric Substitution,

How do I solve this ?





\[\int{\sqrt{{{a}^{2}}+{{x}^{2}}}}\]

Answer 1

Let x= atanu
meaning dx = asec^2(u)
let \[\sqrt{a^{2}+x^{2}} = asecu\]

Answer 2

\[\int\limits_{}^{}asecu * asec^{2}udu \implies a^{2}\int\limits_{}^{}\sec^{3}udu\]
Then go from there.

Answer 3

Didnt wanna assume she didnt know where to go from there xD

Answer 4

Reduction formula is the easiest choice ??

Answer 5

Also can you please explain me this ?


\[\int{\sqrt{{{a}^{2}}+{{x}^{2}}}}=a\sec (u)\]

Answer 6

But okay, this last portion is integration by parts. You have to say that u = secx and v = sec^2(x). Therefore du = sec(x)tan(x) and integralv = tanx. Then using this formula, I plugin values:

\[u \int\limits_{}^{}v - [\int\limits_{}^{}u'*\int\limits_{}^{}v]\]Which gives me:
\[\sec(x)\tan(x) - \int\limits_{}^{}\sec(x)\tan^{2}(x)\]Now we cannot really do much more that would be useful with by parts, but we can use the remaining integral. If I rewrite tan^2(x) as sec^2(x) - 1, I get:
\[\int\limits_{}^{}\sec^{3}xdx = \sec(x)\tan(x) - [\int\limits_{}^{}\sec^{3}x-\int\limits_{}^{}secx]\]We were trying to solve for integral sec^3(x), so that is still there. But now notice I have another sec^(3)x on the right hand side. I can actually combine those as like terms now by adding integral sec^(3)x to both sides:
\[2\int\limits_{}^{}\sec^{3}xdx = secxtanx+\int\limits_{}^{}secxdx\]Integrating secx and then dividing both sides by 2, I solve for integral sec^3(x) and get:
\[\int\limits_{}^{}\sec^{3}u = \frac{ 1 }{ 2 }secutanu +\frac{ 1 }{ 2 }\ln|secu + tanu| + C\]
Now we need to put this back into terms of x ( I know I switched to x partway through, but we are still in terms of u at this point)
At the beginning, we said
x = atanu
sqrt(a^2+x^2) = asecu
this means that tanu = (x/a) and
secu = sqrt(a^2+x^2)/a. Making those substitutions I get:

\[a^{2}(\frac{ x \sqrt{a^{2}+x^{2}} }{ 2a^{2} }+\frac{ 1 }{ 2 }\ln|\frac{ \sqrt{a^{2}+x^{2}} +x}{ a}|) + C \]

As for how I get asec(u), if I say x = atan(u), I get this:
\[\sqrt{a^{2}+(atan(u))^{2}} \implies \sqrt{a^{2}+a^{2}\tan^{2}u}\]factor out a^2
\[\sqrt{a^{2}(1+\tan^{2}u)} = \sqrt{a^{2}\sec^{2}u}= asecu\]

Answer 7

I have to go, though, so hopefully someone can finish up the work for me. Good Luck : )

Answer 8

he is wright