Question 10 of 33intermediate🔧 ApplyShort Answer2 marks

Integrate the functions in Exercises 1 to 37: tan⁡2\tan^2(2x - 3)

Correct Answer

12tan⁡\frac{1}{2} \tan(2x - 3) - x + C

Exercise: EXERCISE 7.2 | Q: 21 | (Page 17)
For More Understanding

Explanation

The question asks for the integral of tan⁡2(2x−3)\tan^2(2x - 3). Although the provided context focuses on integrals like sin⁡3x\sin^3 x and dx1+tan⁡x\frac{dx}{1+\tan x}, it demonstrates the key methods required: simplifying trigonometric identities and using substitution. Following the context's approach of simplifying integrands (like converting sin⁡2x\sin^2 x to 1−cos⁡2x1-\cos^2 x), we use the identity tan⁡2θ=sec⁡2θ−1\tan^2 \theta = \sec^2 \theta - 1. The integral becomes ∫(sec⁡2(2x−3)−1)dx\int (\sec^2(2x-3) - 1) dx. We then split this and apply the substitution method shown in the text for composite functions (like t=cos⁡xt=\cos x or t=x−3t=x-3). Letting t=2x−3t = 2x-3, we solve the integral to get 12tan⁡(2x−3)−x+C\frac{1}{2} \tan(2x-3) - x + C.

Solution Steps

  1. Step 1: Use the trigonometric identity tan⁡2θ=sec⁡2θ−1\tan^2 \theta = \sec^2 \theta - 1 to simplify the integrand: ∫tan⁡2(2x−3)dx=∫(sec⁡2(2x−3)−1)dx\int \tan^2(2x - 3) dx = \int (\sec^2(2x - 3) - 1) dx.

  2. Step 2: Split the integral into two parts: ∫sec⁡2(2x−3)dx−∫1dx\int \sec^2(2x - 3) dx - \int 1 dx.

  3. Step 3: For the first part, apply the substitution method. Let t=2x−3t = 2x - 3. Then dt=2dxdt = 2 dx or dx=dt2dx = \frac{dt}{2}.

  4. Step 4: Substitute into the first integral: ∫sec⁡2(t)dt2=12∫sec⁡2tdt=12tan⁡t=12tan⁡(2x−3)\int \sec^2(t) \frac{dt}{2} = \frac{1}{2} \int \sec^2 t dt = \frac{1}{2} \tan t = \frac{1}{2} \tan(2x - 3).

  5. Step 5: Integrate the second part: −∫1dx=−x-\int 1 dx = -x. Combine the results to get the final answer: 12tan⁡(2x−3)−x+C\frac{1}{2} \tan(2x - 3) - x + C.

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