Question

∫(tanxtan2xtan3x)dx

Answer 1

(1/3)ln|sec3x| - (1/2)ln|sec2x| - ln|secx| + C

Answer 2

To evaluate the integral $\int(\tan x \tan 2x \tan 3x)dx$, we first need to simplify the integrand using a trigonometric identity.

We know that $3x = x + 2x$. Taking the tangent of both sides: $\tan(3x) = \tan(x + 2x)$ Using the tangent addition formula, $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$: $\tan(3x) = \frac{\tan x + \tan 2x}{1 - \tan x \tan 2x}$ Now, rearrange the equation to isolate the product $\tan x \tan 2x \tan 3x$: $\tan(3x)(1 - \tan x \tan 2x) = \tan x + \tan 2x$ $\tan 3x - \tan x \tan 2x \tan 3x = \tan x + \tan 2x$ Move the product term to one side and the other tangent terms to the other side: $\tan x \tan 2x \tan 3x = \tan 3x - \tan x - \tan 2x$ Now, substitute this expression back into the integral: $\int(\tan x \tan 2x \tan 3x)dx = \int(\tan 3x - \tan 2x - \tan x)dx$ We can integrate each term separately. Recall the standard integral formula for tangent: $\int \tan(ax) dx = \frac{1}{a} \ln |\sec(ax)| + C$ Applying this formula to each term: $\int \tan 3x dx = \frac{1}{3} \ln |\sec 3x|$ $\int \tan 2x dx = \frac{1}{2} \ln |\sec 2x|$ $\int \tan x dx = \ln |\sec x|$ Combining these results, the integral becomes: $\int(\tan 3x - \tan 2x - \tan x)dx = \frac{1}{3} \ln |\sec 3x| - \frac{1}{2} \ln |\sec 2x| - \ln |\sec x| + C$