Integrals Area Ed54Db

1. Find the integrals: 1.1 \( \int \cot x \, dx \) Recall that \( \cot x = \frac{\cos x}{\sin x} \) and the integral formula: $$ \int \cot x \, dx = \ln|\sin x| + C $$ 1.2 \( \int \frac{1}{4x^2 + 3} \, dx \) Rewrite denominator as \( 4x^2 + 3 = (2x)^2 + \sqrt{3}^2 \). Use formula for \( \int \frac{dx}{x^2 + a^2} = \frac{1}{a} \arctan \frac{x}{a} + C \). Let \( u = 2x \), then \( du = 2 dx \) or \( dx = \frac{du}{2} \). Integral becomes: $$ \int \frac{1}{u^2 + (\sqrt{3})^2} \cdot \frac{du}{2} = \frac{1}{2} \int \frac{du}{u^2 + (\sqrt{3})^2} = \frac{1}{2 \sqrt{3}} \arctan \frac{u}{\sqrt{3}} + C = \frac{1}{2 \sqrt{3}} \arctan \frac{2x}{\sqrt{3}} + C $$ 1.3 \( \int \frac{x}{2 - x^2} \, dx \) Rewrite denominator as \( 2 - x^2 = -(x^2 - 2) \). Use substitution \( u = 2 - x^2 \), then \( du = -2x \, dx \) or \( -\frac{1}{2} du = x \, dx \). Integral becomes: $$ \int \frac{x}{u} \, dx = \int \frac{-\frac{1}{2} du}{u} = -\frac{1}{2} \int \frac{1}{u} \, du = -\frac{1}{2} \ln|u| + C = -\frac{1}{2} \ln|2 - x^2| + C $$ 1.4 \( \int \frac{1}{2 + 3x^2} \, dx \) Rewrite denominator as \( 2 + 3x^2 = \sqrt{2}^2 + (\sqrt{3}x)^2 \). Use formula \( \int \frac{dx}{a^2 + x^2} = \frac{1}{a} \arctan \frac{x}{a} + C \). Let \( u = \sqrt{3} x \), then \( du = \sqrt{3} dx \) or \( dx = \frac{du}{\sqrt{3}} \). Integral becomes: $$ \int \frac{1}{2 + u^2} \cdot \frac{du}{\sqrt{3}} = \frac{1}{\sqrt{3}} \int \frac{du}{(\sqrt{2})^2 + u^2} = \frac{1}{\sqrt{3} \sqrt{2}} \arctan \frac{u}{\sqrt{2}} + C = \frac{1}{\sqrt{6}} \arctan \frac{\sqrt{3} x}{\sqrt{2}} + C $$ 2. Use substitution \( u = x - 2 \) to find exact value of $$ \int \frac{4}{1 + (x - 2)^2} \, dx $$ Substitute \( u = x - 2 \), so \( du = dx \). Integral becomes: $$ \int \frac{4}{1 + u^2} \, du = 4 \int \frac{1}{1 + u^2} \, du = 4 \arctan u + C = 4 \arctan (x - 2) + C $$ 3. Find exact area under curve \( y = \frac{\pi}{2} \sin^2 x \cos^2 x \) from \( 0 \) to \( \frac{\pi}{2} \) using substitution \( u = \sin x \). Area = $$ \int_0^{\frac{\pi}{2}} \frac{\pi}{2} \sin^2 x \cos^2 x \, dx = \frac{\pi}{2} \int_0^{\frac{\pi}{2}} \sin^2 x \cos^2 x \, dx $$ Substitute \( u = \sin x \), so \( du = \cos x \, dx \). Rewrite \( \cos^2 x = (\cos x)^2 \), so: $$ \sin^2 x \cos^2 x \, dx = u^2 (\cos x)^2 \, dx = u^2 \cos x (\cos x \, dx) = u^2 \cos x \, du $$ But \( \cos x \, dx = du \), so \( \cos^2 x \, dx = \cos x \, du \) is not straightforward. Better to write \( \cos^2 x = 1 - \sin^2 x = 1 - u^2 \). So: $$ \sin^2 x \cos^2 x \, dx = u^2 (1 - u^2) \, dx $$ Since \( du = \cos x \, dx \), then \( dx = \frac{du}{\cos x} = \frac{du}{\sqrt{1 - u^2}} \). But this complicates. Alternatively, use \( \sin x = u \), \( dx = \frac{du}{\cos x} = \frac{du}{\sqrt{1 - u^2}} \). Rewrite integral: $$ \int_0^{\frac{\pi}{2}} \sin^2 x \cos^2 x \, dx = \int_0^1 u^2 (1 - u^2) \frac{du}{\sqrt{1 - u^2}} = \int_0^1 u^2 (1 - u^2) (1 - u^2)^{-1/2} du = \int_0^1 u^2 (1 - u^2)^{1/2} du $$ So area = $$ \frac{\pi}{2} \int_0^1 u^2 (1 - u^2)^{1/2} du $$ Use substitution \( u = \sin \theta \), or use Beta function or integral tables. Use substitution \( u = \sqrt{t} \), or better use Beta function: $$ \int_0^1 u^m (1 - u^n)^p du = \text{Beta function form} $$ Here: $$ \int_0^1 u^2 (1 - u^2)^{1/2} du = \int_0^1 u^2 (1 - u^2)^{1/2} du $$ Let \( t = u^2 \), then \( dt = 2u du \), so \( u du = \frac{dt}{2} \). Rewrite integral: $$ \int_0^1 u^2 (1 - u^2)^{1/2} du = \int_0^1 u^2 (1 - t)^{1/2} du $$ But \( u^2 du = u \cdot u du = u \cdot \frac{dt}{2} = \frac{u dt}{2} \), still complicated. Alternatively, use Beta function: $$ \int_0^1 x^{p-1} (1 - x)^{q-1} dx = \mathrm{B}(p,q) = \frac{\Gamma(p) \Gamma(q)}{\Gamma(p+q)} $$ Rewrite integral as: $$ \int_0^1 u^2 (1 - u^2)^{1/2} du = \int_0^1 u^2 (1 - u^2)^{1/2} du $$ Let \( w = u^2 \), then \( dw = 2u du \), so \( u du = \frac{dw}{2} \). Rewrite integral: $$ \int_0^1 u^2 (1 - u^2)^{1/2} du = \int_0^1 u^2 (1 - w)^{1/2} du $$ But \( u^2 du = u \cdot u du = u \cdot \frac{dw}{2} = \frac{u dw}{2} \), still complicated. Better to use direct Beta function substitution: Set \( t = u^2 \), then \( u = t^{1/2} \), \( du = \frac{1}{2} t^{-1/2} dt \). So: $$ \int_0^1 u^2 (1 - u^2)^{1/2} du = \int_0^1 t^{1} (1 - t)^{1/2} \frac{1}{2} t^{-1/2} dt = \frac{1}{2} \int_0^1 t^{1 - \frac{1}{2}} (1 - t)^{1/2} dt = \frac{1}{2} \int_0^1 t^{\frac{1}{2}} (1 - t)^{\frac{1}{2}} dt $$ This is Beta function \( \mathrm{B}(p,q) \) with \( p = \frac{3}{2} \), \( q = \frac{3}{2} \). So: $$ \int_0^1 t^{\frac{1}{2}} (1 - t)^{\frac{1}{2}} dt = \mathrm{B}\left(\frac{3}{2}, \frac{3}{2}\right) = \frac{\Gamma\left(\frac{3}{2}\right)^2}{\Gamma(3)} $$ Recall: $$ \Gamma\left(\frac{3}{2}\right) = \frac{1}{2} \sqrt{\pi} $$ and $$ \Gamma(3) = 2! = 2 $$ Therefore: $$ \mathrm{B}\left(\frac{3}{2}, \frac{3}{2}\right) = \frac{\left(\frac{1}{2} \sqrt{\pi}\right)^2}{2} = \frac{\frac{1}{4} \pi}{2} = \frac{\pi}{8} $$ So the integral is: $$ \frac{1}{2} \cdot \frac{\pi}{8} = \frac{\pi}{16} $$ Finally, area = $$ \frac{\pi}{2} \times \frac{\pi}{16} = \frac{\pi^2}{32} $$ **Final answers:** 1.1 \( \int \cot x \, dx = \ln|\sin x| + C \) 1.2 \( \int \frac{1}{4x^2 + 3} \, dx = \frac{1}{2 \sqrt{3}} \arctan \frac{2x}{\sqrt{3}} + C \) 1.3 \( \int \frac{x}{2 - x^2} \, dx = -\frac{1}{2} \ln|2 - x^2| + C \) 1.4 \( \int \frac{1}{2 + 3x^2} \, dx = \frac{1}{\sqrt{6}} \arctan \frac{\sqrt{3} x}{\sqrt{2}} + C \) 2. \( \int \frac{4}{1 + (x - 2)^2} \, dx = 4 \arctan (x - 2) + C \) 3. Area under \( y = \frac{\pi}{2} \sin^2 x \cos^2 x \) from 0 to \( \frac{\pi}{2} \) is \( \frac{\pi^2}{32} \).