Question

$\int \frac{\sec x}{3(\sec x + \tan x)+2}dx =$

Answer 1

$\frac{1}{2} \ln\left|\frac{\tan(x/2)+1}{\tan(x/2)+5}\right| + C$

Answer 2

The given integral is $I = \int \frac{\sec x}{3(\sec x + \tan x)+2}dx$.

First, rewrite the integrand in terms of $\sin x$ and $\cos x$: $\sec x = \frac{1}{\cos x}$ and $\tan x = \frac{\sin x}{\cos x}$. The integrand becomes: $\frac{\frac{1}{\cos x}}{3(\frac{1}{\cos x} + \frac{\sin x}{\cos x})+2} = \frac{\frac{1}{\cos x}}{3(\frac{1+\sin x}{\cos x})+2} = \frac{\frac{1}{\cos x}}{\frac{3(1+\sin x) + 2\cos x}{\cos x}}$ $= \frac{1}{\cos x} \times \frac{\cos x}{3(1+\sin x) + 2\cos x} = \frac{1}{3+3\sin x + 2\cos x}$. So the integral is $I = \int \frac{1}{3+3\sin x + 2\cos x}dx$.

This is an integral of the form $\int \frac{1}{a+b\sin x + c\cos x}dx$. We use the substitution $t = \tan(x/2)$. We have $dx = \frac{2dt}{1+t^2}$, $\sin x = \frac{2t}{1+t^2}$, and $\cos x = \frac{1-t^2}{1+t^2}$. Substitute these into the integral: $I = \int \frac{1}{3+3\left(\frac{2t}{1+t^2}\right) + 2\left(\frac{1-t^2}{1+t^2}\right)} \frac{2dt}{1+t^2}$ Simplify the denominator: $3+3\left(\frac{2t}{1+t^2}\right) + 2\left(\frac{1-t^2}{1+t^2}\right) = \frac{3(1+t^2) + 3(2t) + 2(1-t^2)}{1+t^2} = \frac{3+3t^2+6t+2-2t^2}{1+t^2} = \frac{t^2+6t+5}{1+t^2}$. Substitute this back into the integral: $I = \int \frac{1}{\frac{t^2+6t+5}{1+t^2}} \frac{2dt}{1+t^2} = \int \frac{1+t^2}{t^2+6t+5} \frac{2dt}{1+t^2}$ $I = \int \frac{2}{t^2+6t+5} dt$.

Factor the quadratic in the denominator: $t^2+6t+5 = (t+1)(t+5)$. $I = \int \frac{2}{(t+1)(t+5)} dt$.

Use partial fraction decomposition: $\frac{2}{(t+1)(t+5)} = \frac{A}{t+1} + \frac{B}{t+5}$ $2 = A(t+5) + B(t+1)$. Set $t=-1$: $2 = A(-1+5) + B(-1+1) = 4A \implies A = \frac{1}{2}$. Set $t=-5$: $2 = A(-5+5) + B(-5+1) = -4B \implies B = -\frac{1}{2}$. So, $\frac{2}{(t+1)(t+5)} = \frac{1/2}{t+1} - \frac{1/2}{t+5}$.

Now integrate with respect to $t$: $I = \int \left(\frac{1/2}{t+1} - \frac{1/2}{t+5}\right) dt = \frac{1}{2}\int \frac{1}{t+1} dt - \frac{1}{2}\int \frac{1}{t+5} dt$ $I = \frac{1}{2} \ln|t+1| - \frac{1}{2} \ln|t+5| + C$ $I = \frac{1}{2} (\ln|t+1| - \ln|t+5|) + C$ $I = \frac{1}{2} \ln\left|\frac{t+1}{t+5}\right| + C$.

Substitute back $t = \tan(x/2)$: $I = \frac{1}{2} \ln\left|\frac{\tan(x/2)+1}{\tan(x/2)+5}\right| + C$.