Question

Integration of tan^-1(cos2x-sin2x/cos2x+sin2x))

Answer 1

(π/4)x - x^2 + C

Answer 2

The problem asks for the integration of $\tan^{-1}\left(\frac{\cos 2x - \sin 2x}{\cos 2x + \sin 2x}\right)$.

Let the given integral be $I$. $I = \int \tan^{-1}\left(\frac{\cos 2x - \sin 2x}{\cos 2x + \sin 2x}\right) dx$

Step 1: Simplify the argument of the inverse tangent function.

Divide the numerator and denominator of the fraction inside the $\tan^{-1}$ by $\cos 2x$: $\frac{\cos 2x - \sin 2x}{\cos 2x + \sin 2x} = \frac{\frac{\cos 2x}{\cos 2x} - \frac{\sin 2x}{\cos 2x}}{\frac{\cos 2x}{\cos 2x} + \frac{\sin 2x}{\cos 2x}} = \frac{1 - \tan 2x}{1 + \tan 2x}$ We know the trigonometric identity for $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$. By setting $A = \frac{\pi}{4}$ (since $\tan\left(\frac{\pi}{4}\right) = 1$) and $B = 2x$, we can write: $\frac{1 - \tan 2x}{1 + \tan 2x} = \frac{\tan\left(\frac{\pi}{4}\right) - \tan 2x}{1 + \tan\left(\frac{\pi}{4}\right)\tan 2x} = \tan\left(\frac{\pi}{4} - 2x\right)$ So, the argument of the inverse tangent simplifies to $\tan\left(\frac{\pi}{4} - 2x\right)$.

Step 2: Simplify the integrand using the property of inverse trigonometric functions.

Substitute the simplified argument back into the integral: $I = \int \tan^{-1}\left(\tan\left(\frac{\pi}{4} - 2x\right)\right) dx$ For the principal value branch of $\tan^{-1}$, we have $\tan^{-1}(\tan \theta) = \theta$. Assuming that $\left(\frac{\pi}{4} - 2x\right)$ lies within the range $(-\frac{\pi}{2}, \frac{\pi}{2})$, the integrand simplifies to: $I = \int \left(\frac{\pi}{4} - 2x\right) dx$

Step 3: Perform the integration.

Now, integrate the simplified expression term by term: $I = \int \frac{\pi}{4} dx - \int 2x dx$ $I = \frac{\pi}{4}x - 2\left(\frac{x^{1+1}}{1+1}\right) + C$ $I = \frac{\pi}{4}x - 2\left(\frac{x^2}{2}\right) + C$ $I = \frac{\pi}{4}x - x^2 + C$ where $C$ is the constant of integration.