Question

if we let x=tan theta, then integral of (1+x^2)^(1/2) dx from 1to 3^(1/2) is equivalent to

Answer 1

This is your integral
\[\int\limits_{1}^{\sqrt{3}}\sqrt{1+x^2} dx\]
\[x = \tan \theta \rightarrow dx = \sec^2 \theta ~d \theta\]
\[\sqrt{1+x^2} = \sqrt{1+\tan^2 \theta} = \sqrt{\sec^2 \theta} = \left| \sec \theta \right|\]
Our new limits
\[x = \sqrt{3} \rightarrow 1+x^2 = 1+ 3 = 4\]
\[x = 1 \rightarrow 1+ x^2 = 1+1 = 2\]
Together we can write our integral as
\[\int\limits_{1}^{\sqrt 3} \sqrt{1+x^2}dx = \int\limits_{2}^{4}\left| \sec \theta \right| \sec^2 \theta ~d \theta = \int\limits_{2}^{4} \sec^3 \theta ~d \theta\]

This is a pretty tricky integral too so what you can do is rewrite as follows and use integration by parts, noting recursion of the original integral.

\[Let ~I = \int\limits_{2}^{4} \sec^3 \theta ~d \theta = \int\limits_{2}^{4} \sec \theta \sec^2 \theta ~d \theta = [\tan \theta \sec \theta]_{2}^{4} - \int\limits_{2}^{4} \tan^2 \theta \sec \theta ~d \theta\]
\[I = \left[ \tan \theta \sec \theta \right]_{2}^{4} - \int\limits_{2}^{4} (\sec^2 \theta - 1)\sec \theta ~d \theta = \left[ \tan \theta \sec \theta \right]_{2}^{4} - \int\limits_{2}^{4} (\sec^3 \theta - \sec \theta) ~d \theta \]
\[I = \left[ \tan \theta \sec \theta \right]_{2}^{4} - I + \int\limits_{2}^{4} \sec \theta ~d \theta\]

Tranpose the I on the RHS to the LHS and evaluate the resulting integral. Note the integral of sec is ln|sec x + tan x|

Answer 2

Interval in new variable is \(\pi/4 \to \pi /3\)