Question 22 of 33intermediate🔧 ApplyShort Answer2 marks

Find the integrals of the functions in Exercises 1 to 22: tan⁡3\tan^3 2x sec⁡\sec 2x

Correct Answer

To find the integral of tan⁡32xsec⁡2x\tan^3 2x \sec 2x, we first express the function in terms of sine and cosine. Using the definition tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}, we rewrite the integrand as ∫sin⁡32xcos⁡42x dx\int \frac{\sin^3 2x}{\cos^4 2x} \, dx. We separate one sine factor for the derivative of cosine: ∫sin⁡22xcos⁡42xsin⁡2x dx\int \frac{\sin^2 2x}{\cos^4 2x} \sin 2x \, dx. Using the identity sin⁡22x=1−cos⁡22x\sin^2 2x = 1 - \cos^2 2x, the integral becomes ∫1−cos⁡22xcos⁡42xsin⁡2x dx\int \frac{1 - \cos^2 2x}{\cos^4 2x} \sin 2x \, dx.

Next, we use the substitution method shown in the text by putting cos⁡2x=t\cos 2x = t. Differentiating gives −2sin⁡2x dx=dt-2 \sin 2x \, dx = dt, or sin⁡2x dx=−12dt\sin 2x \, dx = -\frac{1}{2} dt. Substituting these values, the integral transforms into −12∫1−t2t4 dt-\frac{1}{2} \int \frac{1 - t^2}{t^4} \, dt. Simplifying and integrating term by term, we get −12∫(t−4−t−2) dt=−12[t−3−3−t−1−1]+C-\frac{1}{2} \int (t^{-4} - t^{-2}) \, dt = -\frac{1}{2} [\frac{t^{-3}}{-3} - \frac{t^{-1}}{-1}] + C. This simplifies to 16t3−12t+C\frac{1}{6t^3} - \frac{1}{2t} + C.

Finally, substituting t=cos⁡2xt = \cos 2x back into the expression, we get 16cos⁡32x−12cos⁡2x+C\frac{1}{6 \cos^3 2x} - \frac{1}{2 \cos 2x} + C. Writing this in terms of secant, the final answer is 16sec⁡32x−12sec⁡2x+C\frac{1}{6} \mathbf{\sec^3 2x} - \frac{1}{2} \mathbf{\sec 2x} + C.

Exercise: EXERCISE 7.3 | Q: 15 | (Page 19)
For More Understanding

Explanation

The solution follows the standard method of integration by substitution demonstrated in the provided context for integrals involving trigonometric functions like ∫tan⁡x dx\int \tan x \, dx and ∫sin⁡3x dx\int \sin^3 x \, dx. Specifically, it utilizes the technique of expressing tan⁡x\tan x as sin⁡xcos⁡x\frac{\sin x}{\cos x} and using the substitution cos⁡x=t\cos x = t (or cos⁡2x=t\cos 2x = t in this case) to simplify the integral into a polynomial form that is easily integrable.

Solution Steps

  1. Step 1: Rewrite the integrand using tan⁡2x=sin⁡2xcos⁡2x\tan 2x = \frac{\sin 2x}{\cos 2x} and sec⁡2x=1cos⁡2x\sec 2x = \frac{1}{\cos 2x} to get ∫sin⁡32xcos⁡42x dx\int \frac{\sin^3 2x}{\cos^4 2x} \, dx.

  2. Step 2: Use the identity sin⁡22x=1−cos⁡22x\sin^2 2x = 1 - \cos^2 2x to separate terms: ∫1−cos⁡22xcos⁡42xsin⁡2x dx\int \frac{1 - \cos^2 2x}{\cos^4 2x} \sin 2x \, dx.

  3. Step 3: Apply the substitution cos⁡2x=t\cos 2x = t, so −2sin⁡2x dx=dt-2 \sin 2x \, dx = dt or sin⁡2x dx=−12dt\sin 2x \, dx = -\frac{1}{2} dt.

  4. Step 4: Substitute into the integral to get −12∫1−t2t4 dt=−12∫(t−4−t−2) dt-\frac{1}{2} \int \frac{1 - t^2}{t^4} \, dt = -\frac{1}{2} \int (t^{-4} - t^{-2}) \, dt.

  5. Step 5: Integrate to find −12[t−3−3−t−1−1]+C=16t3−12t+C-\frac{1}{2} [\frac{t^{-3}}{-3} - \frac{t^{-1}}{-1}] + C = \frac{1}{6t^3} - \frac{1}{2t} + C.

  6. Step 6: Substitute back t=cos⁡2xt = \cos 2x and convert to secant form: 16sec⁡32x−12sec⁡2x+C\frac{1}{6} \sec^3 2x - \frac{1}{2} \sec 2x + C.

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