Question

\(\int \cos 2 x \ln (1+\tan x) d x\)

Short answer

Question: Find the integral of the given function: $$\int \cos 2x \ln (1+\tan x) dx$$ Answer: This exercise is quite challenging and requires a deep understanding of calculus concepts. Unfortunately, after several steps of simplification, we find out that the resulting integral is beyond the scope of high school level calculus. Therefore, we cannot provide an elementary closed-form antiderivative for the original exercise.

Step-by-step solution

Identifying u and dv

Let's first choose our functions u and dv: $$u = \ln(1+\tan x)$$ $$dv = \cos 2x \, dx$$

Finding du and v

Now we need to find the derivative of u (du) and the integral of dv (v): To find \(du\): $$u = \ln(1 + \tan x)$$ $$\frac{du}{dx} = \frac{d}{dx}\ln(1 + \tan x) = \frac{(1+\tan x)'}{1 + \tan x}$$ We need to find the derivative of \((1 + \tan x)\): $$(1 + \tan x)' = \sec^2 x$$ Hence, $$\frac{du}{dx} = \frac{\sec^2 x}{1 + \tan x}$$ Now, multiply both sides by \(dx\) to find \(du\): $$du = \frac{\sec^2 x}{1 + \tan x} dx$$ To find \(v\): $$\int dv = \int \cos 2x \, dx$$ $$v = \frac{1}{2}\sin 2x + C$$

Applying Integration by parts

Now use the formula for integration by parts: $$\int udv = uv - \int vdu$$ Plug in the expressions for \(u\), \(du\), \(v\), and \(dv\) that we found in Steps 1 and 2: $$\int \cos 2x \ln (1+\tan x) dx = \left(\frac{1}{2}\sin 2x + C\right)\ln(1+\tan x) - \int \left(\frac{1}{2}\sin 2x + C\right)\frac{\sec^2 x}{1 + \tan x} dx$$ Now, let's evaluate the integral.

Simplifying the Integral

Let's simplify the integral expression: $$\int \cos 2x \ln (1+\tan x) dx = \frac{1}{2}\sin 2x \ln(1+\tan x) - \int \frac{\sin 2x}{2} \frac{\sec^2 x}{1 + \tan x} dx$$ Now, substitute \(\sin 2x = 2\sin x \cos x\) and \(\sec^2 x = 1 + \tan^2 x\): $$\int \cos 2x \ln(1+\tan x) dx = \frac{1}{2} \sin 2x \ln(1+\tan x) - \int 2\sin x \cos x \frac{1 + \tan^2 x}{2(1 + \tan x)} dx$$ Next, substitute \(\tan x = \frac{\sin x}{\cos x}\) and simplify: $$\int \cos 2x \ln(1+\tan x) dx = \frac{1}{2} \sin 2x \ln(1+\tan x) - \int \frac{(\sin^2 x + \sin^2 x \cos^2 x)}{(1 + \sin x)(\cos x)} dx$$

Separating the Integral

Separate the integral into two parts: $$\int \cos 2x \ln(1+\tan x) dx = \frac{1}{2} \sin 2x \ln(1+\tan x) - \int \frac{\sin^2 x}{(1 + \sin x)(\cos x)} dx - \int \frac{\sin^2 x \cos^2 x}{(1 + \sin x)(\cos x)} dx$$ Unfortunately, the resulting integrals are quite challenging to solve and are beyond the scope of high school level calculus. The first integral may be solvable using hyperbolic trigonometric identities, but the second one is unlikely to have a closed-form antiderivative. As a result, we can't proceed to find an elementary closed-form antiderivative for the original exercise.