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Chapter 7: Problem 52

Use the approaches discussed in this section to evaluate the following integrals. $$\int_{0}^{\pi / 8} \sqrt{1-\cos 4 x} d x$$

Short Answer

Expert verified
Question: Evaluate the integral of the function $$\int_0^{\pi/8} \sqrt{1-\cos{4x}} dx$$. Answer: The integral of the given function is $$\frac{2-\sqrt{2}}{2}$$.

Step by step solution

01

Use the half-angle formula for cosine

Recall the half-angle formula for cosine: $$\cos{2\theta} = 1-2\sin^2{\theta}$$ Using this, we can rewrite the integrand $$1 - \cos{4x}$$ as follows, by substituting $$\theta = 2x$$: $$1-\cos{4x} = 1 - (1-2\sin^2{2x}) = 2\sin^2{2x}$$ Therefore, the integral becomes: $$\int_0^{\pi/8} \sqrt{1-\cos{4x}} dx = \int_0^{\pi/8} \sqrt{2\sin^2{2x}} dx$$
02

Simplify the integrand and transform the integral with substitution

We have: $$\sqrt{2\sin^2{2x}} = \sqrt{2} \cdot \sin{2x}$$ Now we have the following integral: $$\int_0^{\pi/8} \sqrt{2} \cdot \sin{2x} dx$$ Let's use the substitution method: Let $$u = 2x$$, hence $$\frac{1}{2} du = dx$$ When $$x = 0$$, $$u = 0$$ When $$x = \pi/8$$, $$u = \pi/4$$ Now the integral becomes: $$\int_0^{\pi/4} \sqrt{2} \cdot \sin{u} (\frac{1}{2}du) = \frac{\sqrt{2}}{2} \int_0^{\pi/4} \sin{u} du$$
03

Evaluate the integral and then plug in the limits of integration

Integrate with respect to $$u$$: $$\frac{\sqrt{2}}{2} \int_0^{\pi/4} \sin{u} du = \frac{\sqrt{2}}{2} [-\cos{u}]_0^{\pi/4}$$ Now plug in the limits of integration: $$\frac{\sqrt{2}}{2} [-\cos{(\pi/4)} + \cos{0}] = \frac{\sqrt{2}}{2} [-\frac{\sqrt{2}}{2} + 1] = \frac{\sqrt{2}}{2}(\frac{2-\sqrt{2}}{2}) = \boxed{\frac{2-\sqrt{2}}{2}}$$

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