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Value of int(0)^(pi//2) cos 3t is...

Value of `int_(0)^(pi//2) cos 3t` is

A

`(2)/(3)`

B

`-(1)/(3)`

C

`- (2)/(3)`

D

`(1)/(3)`

Text Solution

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The correct Answer is:
To solve the integral \( \int_{0}^{\frac{\pi}{2}} \cos(3t) \, dt \), we will follow these steps: ### Step 1: Identify the integral We need to evaluate the integral: \[ \int_{0}^{\frac{\pi}{2}} \cos(3t) \, dt \] ### Step 2: Use substitution To solve this integral, we will use the substitution method. Let: \[ u = 3t \quad \Rightarrow \quad du = 3 \, dt \quad \Rightarrow \quad dt = \frac{du}{3} \] ### Step 3: Change the limits of integration When \( t = 0 \): \[ u = 3 \cdot 0 = 0 \] When \( t = \frac{\pi}{2} \): \[ u = 3 \cdot \frac{\pi}{2} = \frac{3\pi}{2} \] ### Step 4: Rewrite the integral Now we can rewrite the integral in terms of \( u \): \[ \int_{0}^{\frac{\pi}{2}} \cos(3t) \, dt = \int_{0}^{\frac{3\pi}{2}} \cos(u) \cdot \frac{du}{3} \] This simplifies to: \[ \frac{1}{3} \int_{0}^{\frac{3\pi}{2}} \cos(u) \, du \] ### Step 5: Evaluate the integral The integral of \( \cos(u) \) is \( \sin(u) \): \[ \frac{1}{3} \left[ \sin(u) \right]_{0}^{\frac{3\pi}{2}} \] Now we evaluate this at the limits: \[ = \frac{1}{3} \left( \sin\left(\frac{3\pi}{2}\right) - \sin(0) \right) \] We know that: \[ \sin\left(\frac{3\pi}{2}\right) = -1 \quad \text{and} \quad \sin(0) = 0 \] Thus: \[ = \frac{1}{3} \left( -1 - 0 \right) = -\frac{1}{3} \] ### Final Answer The value of the integral \( \int_{0}^{\frac{\pi}{2}} \cos(3t) \, dt \) is: \[ -\frac{1}{3} \]

To solve the integral \( \int_{0}^{\frac{\pi}{2}} \cos(3t) \, dt \), we will follow these steps: ### Step 1: Identify the integral We need to evaluate the integral: \[ \int_{0}^{\frac{\pi}{2}} \cos(3t) \, dt \] ...
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