Question: Create a detailed educational Maths video for ICS Part 2 students. Explain MCQs 1–5 one by one in simple Urdu with English mathematical terms. For every MCQ: - Show the complete question and all options. - Explain the relevant formula or rule. - Solve step by step on screen. - Explain why the correct option is right. - Use a teacher-like voice, clear board writing, and pauses between steps. Questions: 1. Find d/dx(cot⁻¹ x). Options: A) 1/(1−x²), B) 1/(1+x²), C) −1/(1+x²), D) −1/(1−x²). Correct: C. 2. Evaluate ∫ sin x cos x dx. Options: A) cos²x/2 + C, B) −sin²x/2 + C, C) sin²x/2 + C, D) tan²x/2 + C. Correct: C. 3. Evaluate ∫ eˣ(cos x + sin x) dx. Options: A) eˣ cos x + C, B) eˣ sin x + C, C) eˣ + C, D) eˣ(cos x − sin x) + C. Correct: B. 4. If y = e⁻ᵃˣ, find dy/dx. Options: A) ae⁻ᵃˣ, B) −e⁻ᵃˣ, C) aeᵃˣ, D) −ae⁻ᵃˣ. Correct: D. 5. Differentiate 1/cosec x. Options: A) cos x, B) −cos x, C) sin x, D) −sin x. Correct: A. End with a quick revision of all five answers. Do not skip any calculation or change the questions.

1. **مسئلہ 1:** Find $\frac{d}{dx}(\cot^{-1} x)$. Options: A) $\frac{1}{1-x^2}$, B) $\frac{1}{1+x^2}$, C) $-\frac{1}{1+x^2}$, D) $-\frac{1}{1-x^2}$. **فارمولا:** The derivative of $\cot^{-1} x$ is given by: $$\frac{d}{dx}(\cot^{-1} x) = -\frac{1}{1+x^2}$$ **حل:** 1. Recall that $\cot^{-1} x$ is the inverse cotangent function. 2. Using the formula for the derivative of inverse cotangent, we get: $$\frac{d}{dx}(\cot^{-1} x) = -\frac{1}{1+x^2}$$ **صحیح جواب:** Option C. 2. **مسئلہ 2:** Evaluate $\int \sin x \cos x \, dx$. Options: A) $\frac{\cos^2 x}{2} + C$, B) $-\frac{\sin^2 x}{2} + C$, C) $\frac{\sin^2 x}{2} + C$, D) $\frac{\tan^2 x}{2} + C$. **فارمولا:** Use the identity: $$\sin x \cos x = \frac{1}{2} \sin 2x$$ **حل:** 1. Rewrite the integral: $$\int \sin x \cos x \, dx = \int \frac{1}{2} \sin 2x \, dx$$ 2. Factor out $\frac{1}{2}$: $$= \frac{1}{2} \int \sin 2x \, dx$$ 3. Integrate $\sin 2x$: $$\int \sin 2x \, dx = -\frac{\cos 2x}{2} + C$$ 4. Substitute back: $$= \frac{1}{2} \left(-\frac{\cos 2x}{2}\right) + C = -\frac{\cos 2x}{4} + C$$ 5. Use the identity: $$\cos 2x = 1 - 2 \sin^2 x$$ 6. Substitute: $$-\frac{\cos 2x}{4} = -\frac{1 - 2 \sin^2 x}{4} = -\frac{1}{4} + \frac{\sin^2 x}{2}$$ 7. Absorb constants into $C$: $$= \frac{\sin^2 x}{2} + C$$ **صحیح جواب:** Option C. 3. **مسئلہ 3:** Evaluate $\int e^x (\cos x + \sin x) \, dx$. Options: A) $e^x \cos x + C$, B) $e^x \sin x + C$, C) $e^x + C$, D) $e^x (\cos x - \sin x) + C$. **فارمولا:** Use integration by parts or recognize the pattern: $$\int e^x (\cos x + \sin x) \, dx$$ **حل:** 1. Let: $$I = \int e^x (\cos x + \sin x) \, dx = \int e^x \cos x \, dx + \int e^x \sin x \, dx$$ 2. Use known integrals: $$\int e^x \cos x \, dx = \frac{e^x (\sin x + \cos x)}{2} + C$$ $$\int e^x \sin x \, dx = \frac{e^x (\sin x - \cos x)}{2} + C$$ 3. Add them: $$I = \frac{e^x (\sin x + \cos x)}{2} + \frac{e^x (\sin x - \cos x)}{2} + C$$ 4. Simplify inside parentheses: $$= \frac{e^x (\sin x + \cos x + \sin x - \cos x)}{2} + C = \frac{e^x (2 \sin x)}{2} + C = e^x \sin x + C$$ **صحیح جواب:** Option B. 4. **مسئلہ 4:** If $y = e^{-ax}$, find $\frac{dy}{dx}$. Options: A) $a e^{-ax}$, B) $-e^{-ax}$, C) $a e^{ax}$, D) $-a e^{-ax}$. **فارمولا:** Derivative of exponential function: $$\frac{d}{dx} e^{u} = e^{u} \frac{du}{dx}$$ **حل:** 1. Here, $u = -ax$, so: $$\frac{du}{dx} = -a$$ 2. Therefore: $$\frac{dy}{dx} = e^{-ax} \times (-a) = -a e^{-ax}$$ **صحیح جواب:** Option D. 5. **مسئلہ 5:** Differentiate $\frac{1}{\csc x}$. Options: A) $\cos x$, B) $-\cos x$, C) $\sin x$, D) $-\sin x$. **فارمولا:** Recall: $$\csc x = \frac{1}{\sin x} \implies \frac{1}{\csc x} = \sin x$$ **حل:** 1. Simplify: $$\frac{1}{\csc x} = \sin x$$ 2. Differentiate: $$\frac{d}{dx} (\sin x) = \cos x$$ **صحیح جواب:** Option A. --- **خلاصہ:** 1. $\frac{d}{dx}(\cot^{-1} x) = -\frac{1}{1+x^2}$ (Option C) 2. $\int \sin x \cos x \, dx = \frac{\sin^2 x}{2} + C$ (Option C) 3. $\int e^x (\cos x + \sin x) \, dx = e^x \sin x + C$ (Option B) 4. $\frac{dy}{dx} (e^{-ax}) = -a e^{-ax}$ (Option D) 5. $\frac{d}{dx} \left( \frac{1}{\csc x} \right) = \cos x$ (Option A)