Question 23 of 33intermediate🔧 ApplyShort Answer2 marks

Find the integrals of the functions in Exercises 1 to 22: tan⁡4\tan^4 x

Correct Answer

13tan⁡3\frac{1}{3} \tan^3 x - tan⁡\tan x + x + C

Exercise: EXERCISE 7.3 | Q: 16 | (Page 19)
For More Understanding

Explanation

The question asks for the integral of tan⁡4x\tan^4 x. Although the specific reduction formula is not explicitly detailed in the provided text chunks, the solution relies on the substitution technique demonstrated in the context (Chunk 2) for ∫tan⁡4tsec⁡2t dt\int \tan^4 t \sec^2 t \, dt and the standard integral of tan⁡x\tan x (Chunk 4). We use the trigonometric identity tan⁡2x=sec⁡2x−1\tan^2 x = \sec^2 x - 1 to reduce the power of the tangent function, a standard method in Class 12 NCERT syllabus, and apply the substitution tan⁡x=t\tan x = t.

Solution Steps

  1. Step 1: Express tan⁡4x\tan^4 x as tan⁡2x⋅tan⁡2x\tan^2 x \cdot \tan^2 x and use the identity tan⁡2x=sec⁡2x−1\tan^2 x = \sec^2 x - 1.

  2. Step 2: Rewrite the integral: ∫tan⁡2x(sec⁡2x−1) dx=∫tan⁡2xsec⁡2x dx−∫tan⁡2x dx\int \tan^2 x (\sec^2 x - 1) \, dx = \int \tan^2 x \sec^2 x \, dx - \int \tan^2 x \, dx.

  3. Step 3: For the first integral, use substitution tan⁡x=t\tan x = t, so sec⁡2x dx=dt\sec^2 x \, dx = dt. This gives ∫t2 dt=t33=tan⁡3x3\int t^2 \, dt = \frac{t^3}{3} = \frac{\tan^3 x}{3}.

  4. Step 4: For the second integral, substitute tan⁡2x=sec⁡2x−1\tan^2 x = \sec^2 x - 1 again: ∫(sec⁡2x−1) dx=tan⁡x−x\int (\sec^2 x - 1) \, dx = \tan x - x.

  5. Step 5: Combine the results to get the final answer: 13tan⁡3x−tan⁡x+x+C\frac{1}{3}\tan^3 x - \tan x + x + C.

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