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JEE Main 2019
Indefinite Integration
Indefinite Integrals
Hard

Question

Let In=tannxdx,(n>1).{I_n} = \int {{{\tan }^n}x\,dx} ,\,\left( {n > 1} \right). If I4+I6{I_4} + {I_6} = atan5x+bx5+Ca{\tan ^5}x + b{x^5} + C, where C is a constant of integration, then the ordered pair (a,b)\left( {a,b} \right) is equal to

Options

Solution

Key Concepts and Formulas

  • Indefinite Integral: An indefinite integral represents the family of functions whose derivative is the integrand. We denote it as f(x)dx=F(x)+C\int f(x) \, dx = F(x) + C, where F(x)=f(x)F'(x) = f(x) and CC is the constant of integration.
  • Trigonometric Identity: 1+tan2x=sec2x1 + \tan^2 x = \sec^2 x
  • Substitution Method: If f(g(x))g(x)dx\int f(g(x))g'(x) \, dx can be simplified by substituting u=g(x)u = g(x), then du=g(x)dxdu = g'(x) \, dx and the integral becomes f(u)du\int f(u) \, du.

Step-by-Step Solution

Step 1: Write down the given information. We are given In=tannxdxI_n = \int \tan^n x \, dx, where n>1n > 1. We are also given that I4+I6=atan5x+bx5+CI_4 + I_6 = a \tan^5 x + bx^5 + C, and we need to find the ordered pair (a,b)(a, b).

Step 2: Express I4I_4 and I6I_6 in terms of integrals. I4=tan4xdxI_4 = \int \tan^4 x \, dx and I6=tan6xdxI_6 = \int \tan^6 x \, dx.

Step 3: Substitute I4I_4 and I6I_6 into the given equation and simplify the integral. We have I4+I6=tan4xdx+tan6xdx=(tan4x+tan6x)dxI_4 + I_6 = \int \tan^4 x \, dx + \int \tan^6 x \, dx = \int (\tan^4 x + \tan^6 x) \, dx. We can factor out tan4x\tan^4 x from the integrand: (tan4x+tan6x)dx=tan4x(1+tan2x)dx\int (\tan^4 x + \tan^6 x) \, dx = \int \tan^4 x (1 + \tan^2 x) \, dx

Step 4: Use the trigonometric identity to simplify the integrand. Since 1+tan2x=sec2x1 + \tan^2 x = \sec^2 x, we have tan4x(1+tan2x)dx=tan4xsec2xdx\int \tan^4 x (1 + \tan^2 x) \, dx = \int \tan^4 x \sec^2 x \, dx

Step 5: Use the substitution method to evaluate the integral. Let t=tanxt = \tan x. Then, dt=sec2xdxdt = \sec^2 x \, dx. Substituting into the integral, we get tan4xsec2xdx=t4dt\int \tan^4 x \sec^2 x \, dx = \int t^4 \, dt

Step 6: Evaluate the integral with respect to tt. t4dt=t55+C\int t^4 \, dt = \frac{t^5}{5} + C

Step 7: Substitute back to express the result in terms of xx. Since t=tanxt = \tan x, we have t55+C=tan5x5+C\frac{t^5}{5} + C = \frac{\tan^5 x}{5} + C

Step 8: Compare the result with the given expression atan5x+bx5+Ca \tan^5 x + bx^5 + C. We have 15tan5x+C=atan5x+bx5+C\frac{1}{5} \tan^5 x + C = a \tan^5 x + bx^5 + C. Comparing the coefficients of tan5x\tan^5 x and x5x^5, we get a=15a = \frac{1}{5} and b=0b = 0.

Step 9: Write the ordered pair (a,b)(a, b). The ordered pair is (a,b)=(15,0)(a, b) = \left(\frac{1}{5}, 0\right).

Common Mistakes & Tips

  • Forgetting the constant of integration: Always remember to add the constant of integration CC when evaluating indefinite integrals.
  • Incorrect trigonometric identities: Ensure you use the correct trigonometric identities to simplify the integrand.
  • Choosing the right substitution: Selecting the appropriate substitution is crucial for simplifying the integral. In this case, t=tanxt = \tan x made the integration straightforward.

Summary

We were given an expression involving integrals of powers of tanx\tan x and asked to find the values of aa and bb when the expression was simplified to the form atan5x+bx5+Ca \tan^5 x + bx^5 + C. By combining the integrals, using a trigonometric identity, and applying the substitution method, we were able to simplify the integral to 15tan5x+C\frac{1}{5} \tan^5 x + C. Comparing this with the given form, we found that a=15a = \frac{1}{5} and b=0b = 0. Therefore, the ordered pair (a,b)(a, b) is (15,0)\left(\frac{1}{5}, 0\right).

Final Answer The final answer is \boxed{\left( {{1 \over 5},0} \right)}, which corresponds to option (A).

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