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JEE Main 2024
Indefinite Integration
Indefinite Integrals
Hard

Question

esecx\int {{e^{\sec x}}} (secxtanxf(x)+secxtanx+sex2x)dx(\sec x\tan xf(x) + \sec x\tan x + se{x^2}x)dx = e secx f(x) + C then a possible choice of f(x) is :-

Options

Solution

Key Concepts and Formulas

  • Differentiation of Integrals: If f(x)dx=F(x)+C\int f(x) dx = F(x) + C, then ddxF(x)=f(x)\frac{d}{dx} F(x) = f(x).
  • Product Rule of Differentiation: ddx[u(x)v(x)]=u(x)v(x)+u(x)v(x)\frac{d}{dx} [u(x)v(x)] = u'(x)v(x) + u(x)v'(x).
  • Derivatives of Trigonometric Functions: ddxsecx=secxtanx\frac{d}{dx} \sec x = \sec x \tan x and ddxtanx=sec2x\frac{d}{dx} \tan x = \sec^2 x.
  • Integrals of Trigonometric Functions: secxtanxdx=secx+C\int \sec x \tan x \, dx = \sec x + C and sec2xdx=tanx+C\int \sec^2 x \, dx = \tan x + C.

Step-by-Step Solution

Step 1: Write down the given equation. We are given that esecx(secxtanxf(x)+secxtanx+sec2x)dx=esecxf(x)+C\int {{e^{\sec x}}(\sec x\tan xf(x) + \sec x\tan x + \sec^2 x)dx} = {e^{\sec x}}f(x) + C

Step 2: Differentiate both sides of the equation with respect to xx. Differentiating both sides with respect to xx, we have: ddx[esecx(secxtanxf(x)+secxtanx+sec2x)dx]=ddx[esecxf(x)+C]\frac{d}{dx} \left[\int {{e^{\sec x}}(\sec x\tan xf(x) + \sec x\tan x + \sec^2 x)dx} \right] = \frac{d}{dx} \left[{e^{\sec x}}f(x) + C\right] esecx(secxtanxf(x)+secxtanx+sec2x)=ddx[esecxf(x)]+ddx[C]e^{\sec x}(\sec x\tan xf(x) + \sec x\tan x + \sec^2 x) = \frac{d}{dx} \left[e^{\sec x}f(x)\right] + \frac{d}{dx} [C] Using the product rule for differentiation on the right-hand side, esecx(secxtanxf(x)+secxtanx+sec2x)=esecxsecxtanxf(x)+esecxf(x)+0e^{\sec x}(\sec x\tan xf(x) + \sec x\tan x + \sec^2 x) = e^{\sec x} \sec x \tan x f(x) + e^{\sec x} f'(x) + 0

Step 3: Simplify the equation. Divide both sides by esecxe^{\sec x}: secxtanxf(x)+secxtanx+sec2x=secxtanxf(x)+f(x)\sec x\tan xf(x) + \sec x\tan x + \sec^2 x = \sec x \tan x f(x) + f'(x) Subtract secxtanxf(x)\sec x \tan x f(x) from both sides: secxtanx+sec2x=f(x)\sec x\tan x + \sec^2 x = f'(x) Therefore, f(x)=secxtanx+sec2xf'(x) = \sec x\tan x + \sec^2 x.

Step 4: Integrate f(x)f'(x) to find f(x)f(x). f(x)=f(x)dx=(secxtanx+sec2x)dxf(x) = \int f'(x) dx = \int (\sec x\tan x + \sec^2 x) dx f(x)=secxtanxdx+sec2xdxf(x) = \int \sec x\tan x \, dx + \int \sec^2 x \, dx Using the standard integrals, f(x)=secx+tanx+Kf(x) = \sec x + \tan x + K where KK is the constant of integration.

Step 5: Compare with the given options. The options are: (A) xsecx+tanx+1/2x \sec x + \tan x + 1/2 (B) secx+xtanx1/2\sec x + x \tan x - 1/2 (C) secxtanx1/2\sec x - \tan x - 1/2 (D) secx+tanx+1/2\sec x + \tan x + 1/2

Comparing f(x)=secx+tanx+Kf(x) = \sec x + \tan x + K with the options, we see that option (D) has the form secx+tanx+constant\sec x + \tan x + \text{constant}. Thus, a possible choice of f(x)f(x) is secx+tanx+1/2\sec x + \tan x + 1/2.

Common Mistakes & Tips

  • Careful Differentiation: Be meticulous when applying the product rule. A small error can lead to an incorrect derivative and subsequent incorrect integration.
  • Remember the Constant of Integration: Always add the constant of integration when finding indefinite integrals. This constant can be crucial when matching the solution to given options.
  • Check Against Options: After finding a general form of f(x)f(x), compare it carefully with the available options. Look for a specific value of the constant of integration that makes your solution match one of the options.

Summary

We started with the given integral equation and differentiated both sides with respect to xx. This allowed us to find an expression for f(x)f'(x). Integrating f(x)f'(x) gave us f(x)=secx+tanx+Kf(x) = \sec x + \tan x + K, where KK is a constant. By comparing this result with the provided options, we concluded that a possible choice for f(x)f(x) is secx+tanx+1/2\sec x + \tan x + 1/2.

Final Answer The final answer is \boxed{\sec x + \tan x + 1/2}, which corresponds to option (D).

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