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Chapter 7: Problem 33

Evaluate the following definite integrals. $$\int_{0}^{\pi / 2} x \cos 2 x d x$$

Short Answer

Expert verified
Question: Evaluate the integral $\int_{0}^{\pi / 2} x \cos 2x dx$. Answer: $\frac{1}{2}$.

Step by step solution

01

Identify u and dv

Let \(u = x\) and \(dv = \cos 2x dx\).
02

Calculate du and v

Compute the derivatives and antiderivatives: $$du = \frac{d}{dx} (x) dx = dx$$ $$v = \int \cos 2x dx$$ To find v, we can use substitution. Let \(z = 2x\), so \(dz = 2dx\). Then we have: $$v = \frac{1}{2} \int \cos z dz = \frac{1}{2} \sin z = \frac{1}{2} \sin 2x$$
03

Apply the integration by parts formula

Using the formula \(\int u dv = uv - \int v du\), we have: $$\int_{0}^{\pi / 2} x \cos 2x dx = \left[\frac{1}{2}x \sin 2x\right]_0^{\pi/2} - \int_{0}^{\pi / 2} \frac{1}{2} \sin 2x dx$$
04

Evaluate the remaining integral

We simplify the right-hand side and integrate the remaining integral: $$\int_{0}^{\pi / 2} x \cos 2x dx = \left[\frac{1}{2}x \sin 2x\right]_0^{\pi/2} - \left[-\frac{1}{4}\cos 2x\right]_0^{\pi/2}$$
05

Evaluate at the bounds and find the final answer

Now, we will evaluate the expression at the bounds \(\pi/2\) and \(0\): $$\left[\frac{1}{2}x \sin 2x\right]_0^{\pi/2} = \frac{1}{2}(\frac{\pi}{2})\sin(\pi) - \frac{1}{2}(0)\sin(0) = 0$$ $$\left[-\frac{1}{4}\cos 2x\right]_0^{\pi/2} = -\frac{1}{4}\cos(\pi) - (-\frac{1}{4}\cos(0)) = -\frac{1}{4}(-1) - (-\frac{1}{4}(1)) = \frac{1}{2}$$ So, the final answer is the sum of the two expressions: $$\int_{0}^{\pi / 2} x \cos 2x dx = 0 + \frac{1}{2} = \boxed{\frac{1}{2}}$$

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