Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Question If the area of the region bounded by the curves y = 4 − x 2 4 and y = x − 2 y = 4-\frac{x^2}{4} \quad\text{and}\quad y = x-2 is equal to α \alpha α , then the value of 6 α 6\alpha is equal to ? 🖼️ Question Image ✍️ Short Solution We will: ✔ Find points of intersection ✔ Write the top – bottom function ✔ Integrate between limits ✔ Multiply result by 6 🔹 Step 1 — Find intersection points Solve 4 − x 2 4 = x − 2 4-\frac{x^2}{4}=x-2 Bring to one side: 4 − x 2 4 − x + 2 = 0 ⇒ 6 − x − x 2 4 = 0 4-\frac{x^2}{4}-x+2=0 \Rightarrow 6-x-\frac{x^2}{4}=0 Multiply by 4: 24 − 4 x − x 2 = 0 24-4x-x^2=0 Rewrite: x 2 + 4 x − 24 = 0 x^2+4x-24=0 Solve: x = − 4 ± 16 + 96 2 = − 4 ± 4 7 2 = − 2 ± 2 7 x=\frac{-4\pm\sqrt{16+96}}{2} =\frac{-4\pm 4\sqrt{7}}{2} =-2\pm 2\sqrt{7} ​ (If you used y = x − 2 y=x-2 y = x − 2 , the algebra simplifies the same way — both limits are correct for the standard statement of this problem.) 🔹 Step 2 — Identify which curve is on top...

  ❓ Concept Integration Trick – tan⁻¹ Form in 60 Sec! Whenever you see an integral of the type ∫ sin ⁡ x a + b cos ⁡ 2 x   d x (or with extra terms in the numerator), understand one thing clearly 👇 👉 This is a pure tan⁻¹ game . ✍️ Short Explanation This type of integral is very common in JEE Main + Advanced . The key idea is: sin x dx is the derivative of cos x The denominator becomes a quadratic in cos x Final answer always involves tan⁻¹ 1️⃣ Step 1 — Identify the Pattern Integral of the form: ∫ (something) ⋅ sin ⁡ x a + b cos ⁡ 2 x   d x 👉 Always try substitution : u = cos ⁡ x u = \cos x d u = − sin ⁡ x   d x du = -\sin x\,dx This instantly simplifies the integral. 2️⃣ Step 2 — Apply Substitution From substitution: sin ⁡ x   d x = − d u \sin x\,dx = -du For definite integrals , limits change: x = 0 ⇒ u = cos ⁡ 0 = 1 x = 0 \Rightarrow u = \cos 0 = 1 x = π ⇒ u = cos ⁡ π = − 1 x = \pi \Rightarrow u = \cos \pi = -1 So the integral converts into: ...

  ❓ 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}\,...