The integral $$\int_0^{\frac{\pi}{2}} \frac{dx}{1+(\tan x)^{\pi e}}$$ presents a challenging evaluation due to the combination of the tangent function and the exponent involving Pi and Euler's number. MarkFL provided a correct solution, confirming the complexity of the integral. The discussion highlights the importance of understanding trigonometric functions and their properties in calculus.
PREREQUISITESMathematics students, calculus instructors, and anyone interested in advanced integral evaluation techniques will benefit from this discussion.
MarkFL said:Here is my solution:
We are given to evaluate:
$$I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\pi e}(x)}\,dx$$
Using the property of definite integrals $$\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$$ and a co-function identity, we may state:
$$I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\cot^{\pi e}(x)}\,dx$$
Adding the two equations, we obtain:
$$2I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\pi e}(x)}+\frac{1}{1+\cot^{\pi e}(x)}\,dx$$
$$2I=\int_0^{\frac{\pi}{2}}\frac{2+\tan^{\pi e}(x)+\cot^{\pi e}(x)}{2+\tan^{\pi e}(x)+\cot^{\pi e}(x)}\,dx=\int_0^{\frac{\pi}{2}}\,dx=\frac{\pi}{2}$$
Hence:
$$I=\frac{\pi}{4}$$