Which of the following are anti-derivatives of f (x) = sinxcosx cos” cos(2x) [and II Il and MIL
The Correct Answer and Explanation is:
To find the antiderivatives of f(x)=sin(x)cos(x)f(x) = \sin(x) \cos(x)f(x)=sin(x)cos(x), we can simplify the expression using trigonometric identities. Here’s how you can do it:
Step 1: Simplifying the Function
The function f(x)=sin(x)cos(x)f(x) = \sin(x) \cos(x)f(x)=sin(x)cos(x) can be rewritten using the double angle identity for sine:sin(2x)=2sin(x)cos(x)\sin(2x) = 2\sin(x)\cos(x)sin(2x)=2sin(x)cos(x)
Therefore, we can express sin(x)cos(x)\sin(x)\cos(x)sin(x)cos(x) as:sin(x)cos(x)=12sin(2x)\sin(x) \cos(x) = \frac{1}{2} \sin(2x)sin(x)cos(x)=21sin(2x)
Now, the function becomes:f(x)=12sin(2x)f(x) = \frac{1}{2} \sin(2x)f(x)=21sin(2x)
Step 2: Finding the Antiderivative
To find the antiderivative of 12sin(2x)\frac{1}{2} \sin(2x)21sin(2x), we can apply the standard antiderivative rule for sine functions, which is:∫sin(kx) dx=−1kcos(kx)\int \sin(kx) \, dx = -\frac{1}{k} \cos(kx)∫sin(kx)dx=−k1cos(kx)
In this case, k=2k = 2k=2, so the antiderivative of sin(2x)\sin(2x)sin(2x) is:−12cos(2x)-\frac{1}{2} \cos(2x)−21cos(2x)
Thus, the antiderivative of 12sin(2x)\frac{1}{2} \sin(2x)21sin(2x) is:12⋅(−12cos(2x))=−14cos(2x)\frac{1}{2} \cdot \left( -\frac{1}{2} \cos(2x) \right) = -\frac{1}{4} \cos(2x)21⋅(−21cos(2x))=−41cos(2x)
Step 3: Adding the Constant of Integration
We always add a constant CCC when finding an indefinite integral. So, the antiderivative is:F(x)=−14cos(2x)+CF(x) = -\frac{1}{4} \cos(2x) + CF(x)=−41cos(2x)+C
Conclusion:
The antiderivative of f(x)=sin(x)cos(x)f(x) = \sin(x) \cos(x)f(x)=sin(x)cos(x) is F(x)=−14cos(2x)+CF(x) = -\frac{1}{4} \cos(2x) + CF(x)=−41cos(2x)+C.
If there are multiple choices to pick from, check if this result is listed or closely matches one of the options.
