Question

Find integral sign 10/(4+x^2) dx using the substitution x=2 tan theta. (Give your answer in terms of x.)

Answer 1

\[\int\limits 10/(4+x^2) dx = 10*\int\limits1/(4+x^2) dx\]
Let \[x=2*\tan \theta\]and \[dx = 2*\sec^2\theta d\theta\]
Then,
\[10*\int\limits 1/(4+4*\tan^2 \theta)*2*\sec^2 \theta d\theta\]
Factor out a 4 in the denominator:\[10*\int\limits 1/(4*(1+\tan^2\theta))*2*\sec^2\theta d\theta\]
Use the trig identity \[1+\tan^2\theta=\sec^2\theta\]Then,
\[10*\int\limits (2*\sec^2\theta)/(4*\sec^2\theta) d\theta\]and
\[10*\int\limits 1/2 d\theta=5*\int d\theta=5\theta\]
Now, solve x=2*tan(theta) for theta and get
\[\theta=\tan^-1(x/2)\]Thus, the answer is
\[5*\tan^-1(x/2) + c\]