Question

Evaluate the definite integral: \[ \int_0^{\frac{\pi}{2}} \frac{\sin x}{1 + \cos^2 x} \, dx \]

Hint:

For integrals involving \( \sin x \) and \( \cos x \), substitution like \( u = \cos x \) helps simplify the integral into standard forms.

Answer 1

Let: \[ I = \int_0^{\frac{\pi}{2}} \frac{\sin x}{1 + \cos^2 x} \, dx \] Use substitution: \( u = \cos x \Rightarrow du = -\sin x \, dx \) Change limits: When \( x = 0 \Rightarrow u = 1 \) When \( x = \frac{\pi}{2} \Rightarrow u = 0 \) \[ I = \int_{1}^{0} \frac{-1}{1 + u^2} \, du = \int_{0}^{1} \frac{1}{1 + u^2} \, du = \tan^{-1} u \Big|_0^1 = \tan^{-1} 1 - \tan^{-1} 0 = \frac{\pi}{4} \] Final Answer: \[ \boxed{\frac{\pi}{4}} \]