Subjects calculus

Integral Calculations B3807A

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1. Problem a) Find $$\int \frac{6}{x^2 + 3} \, dx$$ Formula: $$\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \arctan \frac{x}{a} + C$$ Here, $a^2 = 3$, so $a = \sqrt{3}$. Step 1: Factor out the constant 6: $$\int \frac{6}{x^2 + 3} \, dx = 6 \int \frac{1}{x^2 + (\sqrt{3})^2} \, dx$$ Step 2: Apply the formula: $$6 \times \frac{1}{\sqrt{3}} \arctan \frac{x}{\sqrt{3}} + C = \frac{6}{\sqrt{3}} \arctan \frac{x}{\sqrt{3}} + C$$ Step 3: Simplify the coefficient: $$\frac{6}{\sqrt{3}} = 2 \sqrt{3}$$ Final answer a): $$2 \sqrt{3} \arctan \frac{x}{\sqrt{3}} + C$$ 2. Problem b) Find $$\int \sin \left( \frac{x}{2} \right) dx$$ Formula: $$\int \sin(kx) dx = -\frac{1}{k} \cos(kx) + C$$ Here, $k = \frac{1}{2}$. Step 1: Apply the formula: $$-\frac{1}{\frac{1}{2}} \cos \left( \frac{x}{2} \right) + C = -2 \cos \left( \frac{x}{2} \right) + C$$ Final answer b): $$-2 \cos \left( \frac{x}{2} \right) + C$$ 3. Problem c) Find $$\int \frac{1}{2x - 1} dx$$ Formula: $$\int \frac{1}{ax + b} dx = \frac{1}{a} \ln |ax + b| + C$$ Here, $a=2$, $b=-1$. Step 1: Apply the formula: $$\frac{1}{2} \ln |2x - 1| + C$$ Final answer c): $$\frac{1}{2} \ln |2x - 1| + C$$ 4. Problem d) Find $$\int \frac{x}{2 - x^2} dx$$ Rewrite denominator: $$2 - x^2 = -(x^2 - 2)$$ Step 1: Substitute $u = 2 - x^2$, then $du = -2x dx$ or $-\frac{1}{2} du = x dx$ Step 2: Rewrite integral: $$\int \frac{x}{2 - x^2} dx = \int \frac{x dx}{u} = \int \frac{-\frac{1}{2} du}{u} = -\frac{1}{2} \int \frac{1}{u} du$$ Step 3: Integrate: $$-\frac{1}{2} \ln |u| + C = -\frac{1}{2} \ln |2 - x^2| + C$$ Final answer d): $$-\frac{1}{2} \ln |2 - x^2| + C$$ 5. Problem e) Find $$\int \frac{4x - 10}{x^2 - 5x + 1} dx$$ Step 1: Notice numerator is derivative of denominator: $$\frac{d}{dx} (x^2 - 5x + 1) = 2x - 5$$ Rewrite numerator: $$4x - 10 = 2(2x - 5)$$ Step 2: Split integral: $$\int \frac{4x - 10}{x^2 - 5x + 1} dx = 2 \int \frac{2x - 5}{x^2 - 5x + 1} dx$$ Step 3: Use substitution $u = x^2 - 5x + 1$, $du = (2x - 5) dx$ Step 4: Integral becomes: $$2 \int \frac{du}{u} = 2 \ln |u| + C = 2 \ln |x^2 - 5x + 1| + C$$ Final answer e): $$2 \ln |x^2 - 5x + 1| + C$$ 6. Problem f) Find $$\int \tan 3x \, dx$$ Formula: $$\int \tan(kx) dx = -\frac{1}{k} \ln |\cos(kx)| + C$$ Here, $k=3$. Step 1: Apply formula: $$-\frac{1}{3} \ln |\cos 3x| + C$$ Final answer f): $$-\frac{1}{3} \ln |\cos 3x| + C$$