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JEE Main 2023
Indefinite Integration
Indefinite Integrals
Hard

Question

The integral sec2/3xcosec4/3xdx\int {{\rm{se}}{{\rm{c}}^{{\rm{2/ 3}}}}\,{\rm{x }}\,{\rm{cose}}{{\rm{c}}^{{\rm{4 / 3}}}}{\rm{x \,dx}}} is equal to (Hence C is a constant of integration)

Options

Solution

Key Concepts and Formulas

  • Trigonometric Identities: secx=1cosx\sec x = \frac{1}{\cos x}, cscx=1sinx\csc x = \frac{1}{\sin x}, cotx=cosxsinx=1tanx\cot x = \frac{\cos x}{\sin x} = \frac{1}{\tan x}
  • Differentiation of cotangent: ddx(cotx)=csc2x\frac{d}{dx}(\cot x) = -\csc^2 x
  • Power Rule for Integration: xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C, where n1n \neq -1

Step-by-Step Solution

Step 1: Rewrite the integral using trigonometric identities.

The goal is to express the integrand in terms of sine and cosine to simplify it. I=sec2/3xcsc4/3xdx=1cos2/3x1sin4/3xdxI = \int \sec^{2/3} x \csc^{4/3} x \, dx = \int \frac{1}{\cos^{2/3} x} \cdot \frac{1}{\sin^{4/3} x} \, dx

Step 2: Manipulate the expression to isolate csc2x\csc^2 x and cotx\cot x.

We want to create a form where we can use substitution, specifically involving cotx\cot x and its derivative. To do this, we rewrite the expression to isolate a csc2x\csc^2 x term.

I=1cos2/3x1sin2/3x1sin2/3x1sin2/3xdx=1(cosxsinx)2/31sin2xdxI = \int \frac{1}{\cos^{2/3} x} \cdot \frac{1}{\sin^{2/3} x} \cdot \frac{1}{\sin^{2/3} x} \cdot \frac{1}{\sin^{2/3} x} \, dx = \int \frac{1}{(\cos x \sin x)^{2/3}} \cdot \frac{1}{\sin^2 x} \, dx I=1(cosxsinx)2/3sin2/3xsin4/3xdx=1cot2/3xsin2xdx=csc2xcot2/3xdxI = \int \frac{1}{\left(\frac{\cos x}{\sin x}\right)^{2/3} \sin^{2/3}x \sin^{4/3}x} dx = \int \frac{1}{\cot^{2/3} x \sin^2 x} \, dx = \int \frac{\csc^2 x}{\cot^{2/3} x} \, dx

Step 3: Perform a u-substitution.

Let t=cotxt = \cot x. Then, dt=csc2xdxdt = -\csc^2 x \, dx, so csc2xdx=dt\csc^2 x \, dx = -dt. I=dtt2/3=t2/3dtI = \int \frac{-dt}{t^{2/3}} = -\int t^{-2/3} \, dt

Step 4: Evaluate the integral using the power rule.

I=t(2/3)+1(2/3)+1+C=t1/31/3+C=3t1/3+CI = - \frac{t^{(-2/3) + 1}}{(-2/3) + 1} + C = - \frac{t^{1/3}}{1/3} + C = -3t^{1/3} + C

Step 5: Substitute back for x.

Substitute t=cotxt = \cot x back into the expression.

I=3(cotx)1/3+C=3cotx3+CI = -3(\cot x)^{1/3} + C = -3\sqrt[3]{\cot x} + C

Step 6: Rewrite in terms of tangent.

Since cotx=1tanx\cot x = \frac{1}{\tan x}, we have

I=3(1tanx)1/3+C=3(tanx)1/3+C=3tan1/3x+CI = -3 \left(\frac{1}{\tan x}\right)^{1/3} + C = -3 (\tan x)^{-1/3} + C = -3 \tan^{-1/3} x + C

Common Mistakes & Tips

  • Be careful with the signs during u-substitution.
  • Remember to substitute back to the original variable after integration.
  • When dealing with trigonometric integrals, look for opportunities to use trigonometric identities to simplify the expression.

Summary

We started by rewriting the integral using trigonometric identities to get it in terms of sine and cosine. We then manipulated the expression to isolate csc2x\csc^2 x and cotx\cot x to perform a u-substitution with t=cotxt = \cot x. After integrating and substituting back, we expressed the result in terms of tanx\tan x to match one of the answer choices.

Final Answer

The final answer is \boxed{-3 \tan^{-1/3} x + C}, which corresponds to option (D).

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