Problem 14 - Essential Calculus

Question

Find the integral. $$ \int \frac{4 x+3}{\sqrt{1-x^{2}}} d x $$

Step-by-Step Solution

Answer

The integral of \( \frac{4 x+3}{\sqrt{1-x^{2}}} dx \) is \( -4\sqrt{1-x^{2}} + 3\arcsin(x)+C \)

Step 1: Trigonometric Substitution

Substitute \(x = \sin(\theta)\) and \(dx = \cos(\theta) d\theta\) into the integral, so the integral becomes:\[\int \frac{4\sin(\theta) + 3}{\sqrt{1-\sin^2(\theta)}} \cos(\theta) d\theta\]since \(\sqrt{1-\sin^2(\theta)} = \cos(\theta)\), this simplifies to:\[\int (4\sin(\theta) + 3) d\theta \]

Step 2: Integration

The integral can be split, so we get:\[\int 4\sin(\theta) d\theta + \int 3 d\theta\]These two integrals can be easily calculated, to obtain:\[-4\cos(\theta) + 3\theta + C\]

Step 3: Back-substitution

The last step is to transform the result back to the original variable, \(x\), using the inverse trigonometric function. Since \(x = \sin(\theta)\), it follows that \(\theta = \arcsin(x)\) and \(\cos(\theta) = \sqrt{1-x^2}\). We obtain:\[-4\sqrt{1-x^{2}} + 3\arcsin(x)+C\]