The value of $\int_0^{\pi/2} \frac{\sin^{2025} x}{\sin^{2025} x + \cos^{2025} x} dx$ is:

Let $I = \int_0^{\pi/2} \frac{\sin^{n} x}{\sin^{n} x + \cos^{n} x} dx$. Using Property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$: Replace $x$ with $\pi/2 - x$. Note that $\sin(\pi/2 - x) = \cos x$ and $\cos(\pi/2 - x) = \sin x$. $I = \int_0^{\pi/2} \frac{\cos^{n} x}{\cos^{n} x + \sin^{n} x} dx$. Adding the two expressions for $I$: $2I = \int_0^{\pi/2} \frac{\sin^n x + \cos^n x}{\sin^n x + \cos^n x} dx$ $2I = \int_0^{\pi/2} 1 dx = [x]_0^{\pi/2} = \pi/2$. So, $I = \pi/4$.

Definite Integrals

Easymathematics

The value of $\int_{0}^{π /2} \frac{s i n ^{2025} x}{s i n ^{2025} x + c o s ^{2025} x} d x$ is:

Select the correct option:

$π$

$π /2$

$π /4$

$0$

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About This Question

Subject: mathematics
Chapter: integral calculus
Topic: definite integrals
Difficulty: Easy
Year: 2025

Solution

Correct Answer:

$π /4$

Let $I = \int_{0}^{π /2} \frac{s i n ^{n} x}{s i n ^{n} x + c o s ^{n} x} d x$ . Using Property $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ : Replace $x$ with $π /2 - x$ . Note that $sin (π /2 - x) = cos x$ and $cos (π /2 - x) = sin x$ . $I = \int_{0}^{π /2} \frac{c o s ^{n} x}{c o s ^{n} x + s i n ^{n} x} d x$ .

Adding the two expressions for $I$ : $2 I = \int_{0}^{π /2} \frac{s i n ^{n} x + c o s ^{n} x}{s i n ^{n} x + c o s ^{n} x} d x$ $2 I = \int_{0}^{π /2} 1 d x = [x]_{0}^{π /2} = π /2$ . So, $I = π /4$ .

This easy difficulty mathematics question is from the chapter integral calculus, covering the topic of definite integrals. It appeared in the 2025 exam.

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