Question 13 of 33intermediate🔧 ApplyShort Answer2 marks

Find the integrals of the functions in Exercises 1 to 22: sin⁡2\sin^2(2x + 5)

Correct Answer

To find the integral of sin⁡2(2x+5)\sin^2(2x + 5), we first simplify the expression using trigonometric identities. Using the relation sin⁡2θ=1−cos⁡2θ\sin^2 \theta = 1 - \cos^2 \theta and the identity cos⁡2x=1+cos⁡2x2\cos^2 x = \frac{1 + \cos 2x}{2} given in the context, we derive sin⁡2(2x+5)=1−cos⁡(4x+10)2\sin^2(2x + 5) = \frac{1 - \cos(4x + 10)}{2}.

Next, we integrate the simplified expression term by term. The integral becomes 12∫(1−cos⁡(4x+10))dx\frac{1}{2} \int (1 - \cos(4x + 10)) dx, which splits into 12∫1dx−12∫cos⁡(4x+10)dx\frac{1}{2} \int 1 dx - \frac{1}{2} \int \cos(4x + 10) dx. The first term integrates to x2\frac{x}{2}. For the second term, we use substitution, yielding 12⋅14sin⁡(4x+10)\frac{1}{2} \cdot \frac{1}{4} \sin(4x + 10).

Combining these results, the final answer is x2−18sin⁡(4x+10)+C\frac{x}{2} - \frac{1}{8} \sin(4x + 10) + C. This method effectively reduces the power of the sine function to make it integrable.

Exercise: EXERCISE 7.3 | Q: 1 | (Page 19)
For More Understanding

Explanation

The question requires the integration of sin⁡2(2x+5)\sin^2(2x + 5). The provided context in part (i) explicitly solves ∫cos⁡2xdx\int \cos^2 x dx using the identity cos⁡2x=1+cos⁡2x2\cos^2 x = \frac{1 + \cos 2x}{2}. To solve for sin⁡2(2x+5)\sin^2(2x+5), one must use sin⁡2θ=1−cos⁡2θ\sin^2 \theta = 1 - \cos^2 \theta (referenced in the alternative method of context iii) combined with the cosine identity from context (i). This allows the conversion of sin⁡2(2x+5)\sin^2(2x+5) into 1−cos⁡(4x+10)2\frac{1 - \cos(4x+10)}{2}. The subsequent integration follows standard procedures for linear trigonometric arguments as seen throughout the text.

Solution Steps

  1. Step 1: Use the identity cos⁡2x=1+cos⁡2x2\cos^2 x = \frac{1 + \cos 2x}{2} and sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x to write sin⁡2(2x+5)=1−cos⁡(4x+10)2\sin^2(2x + 5) = \frac{1 - \cos(4x + 10)}{2}.

  2. Step 2: Set up the integral: ∫sin⁡2(2x+5)dx=12∫(1−cos⁡(4x+10))dx\int \sin^2(2x + 5) dx = \frac{1}{2} \int (1 - \cos(4x + 10)) dx.

  3. Step 3: Integrate term by term: 12[∫1dx−∫cos⁡(4x+10)dx]\frac{1}{2} [ \int 1 dx - \int \cos(4x + 10) dx ].

  4. Step 4: Apply the integral formula: 12[x−14sin⁡(4x+10)]+C\frac{1}{2} [ x - \frac{1}{4} \sin(4x + 10) ] + C.

  5. Step 5: Final result: x2−18sin⁡(4x+10)+C\frac{x}{2} - \frac{1}{8} \sin(4x + 10) + C.

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