Question

The integral ∫ (2x - 1) cos√((2x - 1)² + 5) / √(4x² - 4x + 6) dx is equal to : (where c is a constant of integration)

• A. (1/2) sin√((2x + 1)² + 5) + c
• B. (1/2) sin√((2x - 1)² + 5) + c
• C. (1/2) cos√((2x - 1)² + 5) + c
• D. (1/2) cos√((2x + 1)² + 5) + c

Hint:

When an integral contains a function $f(g(x))$ multiplied by something that looks like $g'(x)$, use $u$-substitution with $u = g(x)$.

Answer 1

The Correct Option is B

Step 1: Let $u = \sqrt{(2x-1)^2 + 5}$. Note that $4x^2 - 4x + 6 = (2x-1)^2 + 5$.
Step 2: Differentiate $u$: $\frac{du}{dx} = \frac{1}{2\sqrt{(2x-1)^2 + 5}} \cdot 2(2x-1) \cdot 2 = \frac{2(2x-1)}{\sqrt{(2x-1)^2 + 5}}$.
Step 3: Thus, $du = \frac{2(2x-1)}{\sqrt{(2x-1)^2 + 5}} dx$.
Step 4: The integral becomes $\int \frac{1}{2} \cos(u) du = \frac{1}{2} \sin(u) + c$.
Step 5: Substitute $u$ back: $\frac{1}{2} \sin \sqrt{(2x-1)^2 + 5} + c$.