Tough Calculus Question? Try This Shortcut for the π–Limit Integral! ⚡
❓ Question Evaluate the integral: ∫ 0 π ( x + 3 ) sin x 1 + 3 cos 2 x d x 🖼️ Question Image ✍️ Short Solution We split the integral into two parts: I = ∫ 0 π x sin x 1 + 3 cos 2 x d x + 3 ∫ 0 π sin x 1 + 3 cos 2 x d x . I = \int_{0}^{\pi} \frac{x\sin x}{1+3\cos^{2}x}\,dx + 3\int_{0}^{\pi} \frac{\sin x}{1+3\cos^{2}x}\,dx. Let: I 1 = ∫ 0 π x sin x 1 + 3 cos 2 x d x , I 2 = ∫ 0 π sin x 1 + 3 cos 2 x d x . I_1=\int_0^\pi \frac{x\sin x}{1+3\cos^2x}\, dx,\qquad I_2=\int_0^\pi \frac{\sin x}{1+3\cos^2x}\, dx. 🔹 Step 1 — Simplify I 1 I_1 using substitution symmetry Let J = ∫ 0 π x sin x 1 + 3 cos 2 x d x . J=\int_0^\pi \frac{x\sin x}{1+3\cos^2x}\, dx. Substitute: x = π − t , d x = − d t , sin ( π − t ) = sin t , cos ( π − t ) = − cos t . x=\pi - t,\qquad dx=-dt,\qquad \sin(\pi - t)=\sin t,\qquad \cos(\pi - t)=-\cos t. Thus: J = ∫ 0 π ( π − t ) sin t 1 + 3 cos 2 t d t . J=\int_0^\pi \frac{(\pi - t)\sin t}{1+3\cos^2 t}\,...