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Which expression is equal to \[ \int sin^2(2x)dx \]? (A) \( \frac{1}{2} \left( x - \frac{1}{4}sin(4x) \right) + c \) (B) \( \frac{1}{2} \left( x + \frac{1}{4}sin(4x) \right) + c \) (C) \( sin^3(2x) \) + c (D) \( -\frac{cos^2(2x)}{6} \) + c - HSC - SSCE Mathematics Extension 1 - Question 5 - 2016 - Paper 1
Question 5
![Which-expression-is-equal-to-\[-\int-sin^2(2x)dx-\]?---(A)-\(-\frac{1}{2}-\left(-x---\frac{1}{4}sin(4x)-\right)-+-c-\)---(B)-\(-\frac{1}{2}-\left(-x-+-\frac{1}{4}sin(4x)-\right)-+-c-\)---(C)-\(-sin^3(2x)-\)-+-c---(D)-\(--\frac{cos^2(2x)}{6}-\)-+-c-HSC-SSCE Mathematics Extension 1-Question 5-2016-Paper 1.png](https://cdn.simplestudy.cloud/assets/backend/uploads/question/subject_1095_paper_15704_q5_67b626dc.jpg)
Which expression is equal to \[ \int sin^2(2x)dx \]? (A) \( \frac{1}{2} \left( x - \frac{1}{4}sin(4x) \right) + c \) (B) \( \frac{1}{2} \left( x + \frac{1}{4}sin... show full transcript
Worked Solution & Example Answer:Which expression is equal to \[ \int sin^2(2x)dx \]? (A) \( \frac{1}{2} \left( x - \frac{1}{4}sin(4x) \right) + c \) (B) \( \frac{1}{2} \left( x + \frac{1}{4}sin(4x) \right) + c \) (C) \( sin^3(2x) \) + c (D) \( -\frac{cos^2(2x)}{6} \) + c - HSC - SSCE Mathematics Extension 1 - Question 5 - 2016 - Paper 1
Step 1
Determine the integral \( \int sin^2(2x)dx \)
Answer
To solve ( \int sin^2(2x)dx ), we can use the identity ( sin^2(u) = \frac{1 - cos(2u)}{2} ). Applying this identity, we can write: [ \int sin^2(2x)dx = \int \frac{1 - cos(4x)}{2}dx ] This can be split into two parts: [ \frac{1}{2} \int dx - \frac{1}{2} \int cos(4x)dx ]
Step 2
Evaluate the integrals
Answer
The first integral is straightforward: [ \frac{1}{2} \int dx = \frac{x}{2} + C_1 ] For the second integral, we have: [ -\frac{1}{2} \int cos(4x)dx = -\frac{1}{2} \frac{1}{4}sin(4x) + C_2 = -\frac{1}{8}sin(4x) + C_2 ] Combining these results, we get: [ \int sin^2(2x)dx = \frac{x}{2} - \frac{1}{8}sin(4x) + C ]
Step 3