Question

$\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\cot^{4}x}dx=$

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{4}$
• C. $\pi$
• D. $\frac{\pi}{8}$
• E. $2\pi$

Hint:

Integration Tip: Any definite integral of the form $\int_0^{\pi/2} \frac{1}{1+\tan^n x} dx$ or $\int_0^{\pi/2} \frac{1}{1+\cot^n x} dx$ will ALWAYS evaluate to exactly $\frac{\pi}{4}$, regardless of what the power $n$ is!

Answer 1

The Correct Option is B

Concept:
A powerful property of definite integrals is $\int_0^a f(x)dx = \int_0^a f(a-x)dx$. By applying this to trigonometric functions, we can often add the original integral to the transformed integral to cancel out complex denominators perfectly.

Step 1: Define the initial integral.
Let the given integral be $I$: $$I = \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot^{4}x} dx$$

Step 2: Apply the definite integral property.
Substitute $x$ with $(\frac{\pi}{2} - x)$: $$I = \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot^{4}(\frac{\pi}{2}-x)} dx$$

Step 3: Use trigonometric identities.
Since $\cot(\frac{\pi}{2} - x) = \tan x$, the integral becomes: $$I = \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan^{4}x} dx$$ Rewrite $\tan^4 x$ as $\frac{1}{\cot^4 x}$ to find a common denominator: $$I = \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\frac{1}{\cot^{4}x}} dx = \int_{0}^{\frac{\pi}{2}} \frac{\cot^{4}x}{\cot^{4}x+1} dx$$

Step 4: Add the two equations for I.
Add the original integral from Step 1 to the new integral from
Step 3: $$I + I = \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot^{4}x} dx + \int_{0}^{\frac{\pi}{2}} \frac{\cot^{4}x}{1+\cot^{4}x} dx$$ $$2I = \int_{0}^{\frac{\pi}{2}} \frac{1+\cot^{4}x}{1+\cot^{4}x} dx$$

Step 5: Evaluate the simplified integral.
The integrand simplifies entirely to 1: $$2I = \int_{0}^{\frac{\pi}{2}} 1 dx$$ $$2I = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} - 0$$ Divide by 2 to solve for $I$: $$I = \frac{\pi}{4}$$ Hence the correct answer is (B) $\frac{\pi{4}$}.