Question 17 of 33intermediate🔧 ApplyShort Answer2 marks

Find the integrals of the functions in Exercises 1 to 22: sin⁡3\sin^3 x cos⁡3\cos^3 x

Correct Answer

cos⁡6x6−cos⁡4x4\frac{\cos^6 x}{6} - \frac{\cos^4 x}{4} + C

Exercise: EXERCISE 7.3 | Q: 5 | (Page 19)
For More Understanding

Explanation

The question requires integrating a trigonometric function with odd powers. The provided context demonstrates the method for solving integrals of the form ∫sin⁡3xcos⁡2x dx\int \sin^3 x \cos^2 x \, dx by separating one sin⁡x\sin x term and substituting t=cos⁡xt = \cos x. We apply this same method to ∫sin⁡3xcos⁡3x dx\int \sin^3 x \cos^3 x \, dx. By expressing sin⁡3x\sin^3 x as (1−cos⁡2x)sin⁡x(1-\cos^2 x)\sin x, the integral becomes suitable for the substitution t=cos⁡xt = \cos x, which simplifies the trigonometric integral into a simple polynomial integral.

Solution Steps

  1. Step 1: Rewrite the integral using sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x:

    ∫sin⁡3xcos⁡3x dx=∫sin⁡2xcos⁡3xsin⁡x dx=∫(1−cos⁡2x)cos⁡3xsin⁡x dx\int \sin^3 x \cos^3 x \, dx = \int \sin^2 x \cos^3 x \sin x \, dx = \int (1 - \cos^2 x) \cos^3 x \sin x \, dx

  2. Step 2: Apply the substitution Let cos⁡x=t\cos x = t. Then, −sin⁡x dx=dt-\sin x \, dx = dt or sin⁡x dx=−dt\sin x \, dx = -dt. Substituting these into the integral:

    =∫(1−t2)t3(−dt)=−∫(t3−t5) dt= \int (1 - t^2) t^3 (-dt) = -\int (t^3 - t^5) \, dt

  3. Step 3: Integrate the polynomial with respect to tt:

    =−[t44−t66]+C=t66−t44+C= -\left[ \frac{t^4}{4} - \frac{t^6}{6} \right] + C = \frac{t^6}{6} - \frac{t^4}{4} + C

  4. Step 4: Substitute t=cos⁡xt = \cos x back to get the final answer:

    =cos⁡6x6−cos⁡4x4+C= \frac{\cos^6 x}{6} - \frac{\cos^4 x}{4} + C
PreviousNext