Question:
\( \int_{0}^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} \, dx \) is equal to
\( \int_{0}^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} \, dx \) is equal to
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Try $x \to a-x$ trick for symmetric definite integrals.
Updated On: Apr 23, 2026
- $0$
- $\frac{\pi}{4}$
- $\frac{\pi}{2}$
- $\pi$
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The Correct Option is A
Solution and Explanation
Concept:
Use property:
\[
\int_0^{a} f(x)\,dx = \int_0^{a} f(a-x)\,dx
\]
Step 1: Let $I = \int_0^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} dx$.
Step 2: Replace $x \to \frac{\pi}{2} - x$.
\[ I = \int_0^{\pi/2} \frac{\cos x - \sin x}{1 + \sin x \cos x} dx \]
Step 3: Add both expressions.
\[ 2I = \int_0^{\pi/2} 0 \, dx = 0 \]
Step 4: Solve.
\[ I = 0 \] Conclusion:
Answer = $0$
Step 1: Let $I = \int_0^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} dx$.
Step 2: Replace $x \to \frac{\pi}{2} - x$.
\[ I = \int_0^{\pi/2} \frac{\cos x - \sin x}{1 + \sin x \cos x} dx \]
Step 3: Add both expressions.
\[ 2I = \int_0^{\pi/2} 0 \, dx = 0 \]
Step 4: Solve.
\[ I = 0 \] Conclusion:
Answer = $0$
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