Question

Calculate the following integral:

Answer 1

\[\int\limits_{}^{}\sqrt{x^2+1} dx\]

Answer 2

radical will go if you make the substitution \(x=\tan(\theta)\) although there might be an easier way

Answer 3

no, probably not
you get \[\int sec^3(\theta)d\theta\] and that you can look up

Answer 4

it looks good to me, how are you saying it is wrong ?

Answer 5

@satellite73:knew that:
\[\sec^2(x)=\tan^2(x)+1\]
and he's right but when you substitute you'll see the truth:
\[\int\limits_{}^{}\sqrt{x^2+1}[dx]=\int\limits_{}^{}\sqrt{\tan^2(\Theta)+1}[d(\tan(\Theta))]\]
look at the dx's we need derivative there

Answer 6

If x = Tan(Theta)

dx = sec^2 (Theta) d(Theta)

Answer 7

\[\int\limits_{}^{}\sqrt{x^2+1}[dx]=\int\limits_{}^{}\sqrt{\tan^2(\Theta)+1}[d(\tan(\Theta))] \\ = \int \sqrt{\sec^2(\Theta)} \sec^2(\Theta) d\Theta = \int \sec^3\Theta d\Theta \]

Answer 8

Am I missing something here ?

Answer 9

Nope. You are right on the money, Ganeshie8.

Ifrimpanainte, when we do a substitution, you are correct that we need to substitute in for dx. You got arrived at d(Tan(Theta)), but we can break this down further by actually taking the derivative of Tan(Theta) with respect to Theta.

\[\frac{d[Tan(\Theta)]}{d \Theta} = \sec^2(\Theta)\]
\[=> d[Tan(\Theta)] = \sec^2(\Theta) d \Theta\]

Answer 10

Well then why wolframalpha doesn't agree with you guys:
http://www.wolframalpha.com/input/?i=integral&a=*C.integral-_*Calculator.dflt-&f2=(x%5E2%2B1)%5E(1%2F2)&f=Integral.integrand_(x%5E2%2B1)%5E(1%2F2)&a=*FVarOpt.1-_**-.***Integral.rangestart-.*Integral.rangeend--.**Integral.variable---.*--

http://www.wolframalpha.com/input/?i=integral&a=*C.integral-_*Calculator.dflt-&f2=(sec(x))%5E3&f=Integral.integrand_(sec(x))%5E3&a=*FVarOpt.1-_**-.***Integral.rangestart-.*Integral.rangeend--.**Integral.variable---.*--

Answer 11

You haven't completed it. It's far too early to tell whether Wolfram Alpha agree with us or not (It does)

Answer 12

Okay now I'm stucked at this point:
\[\int\limits_{}^{}\sec^3(\Theta)d(\Theta)\]

Answer 13

http://en.wikipedia.org/wiki/Integral_of_secant_cubed

Answer 14

@mathmale