Question

Evaluate the following integrals: $$ \int \frac{x^{2} d x}{\sqrt{1-2 x-x^{2}}} $$

Short answer

Based on the given step-by step solution, create a short answer: To evaluate the integral \(\int \frac{x^{2} d x}{\sqrt{1-2 x-x^{2}}}\), we first rewrite the expression inside the square root by completing the square to get \(\int \frac{x^{2}}{\sqrt{1-(x+1)^{2}}} dx\). We then choose a trigonometric substitution, \(x + 1 = \cos{θ}\), and substitute it into the integral. After simplifying the integral and evaluating it, we obtain a result in terms of θ. Finally, we convert the result back to terms of x, yielding a final answer of \(-\frac{3}{2}\arccos(x+1)-\frac{1}{2}(x+1)\sqrt{1-x^2-2x}+2\sqrt{1-x^2-2x}+C\).

Step-by-step solution

Rewrite the expression inside the square root in a more convenient form

To rewrite the expression inside the square root, we complete the square: $$1-2x-x^2 = -(x^2+2x-1)$$ Now we complete the square for the quadratic inside the parentheses. We take half of the coefficient of the linear term (which is 1) and square it. Then, we add and subtract this number to the quadratic: $$-(x^2+2x+\color{red}{1} -\color{red}{1}-1) = -(x+1)^2+1$$ Now, we can rewrite the integral as: $$ \int \frac{x^{2}}{\sqrt{1-(x+1)^{2}}} dx $$

Choose an appropriate trigonometric substitution

In this problem, we will use a trigonometric substitution to simplify the expression inside the square root. To do that, we will use the identity \(\sin^2{θ} + \cos^2{θ} = 1\). We also want to ensure that the \(dx\) term from the numerator can be replaced with the corresponding \(dθ\) term. The substitution we will use is: $$x + 1 = \cos{θ}$$ Thus, \(dx = -\sin{θ} dθ\). Now, we substitute these expressions back into the integral: $$\int \frac{x^{2}}{\sqrt{1-(x+1)^{2}}} dx = \int \frac{(\cos{θ} - 1)^{2}}{\sqrt{1-\cos^{2}{θ}}} (-\sin{θ}) dθ$$

Simplify and evaluate the integral

We can simplify the expression under the square root using the identity \(\sin^2{θ}+\cos^2{θ}=1\): $$\int \frac{(\cos{θ} - 1)^{2}}{\sqrt{1-\cos^{2}{θ}}} (-\sin{θ}) dθ = \int \frac{(\cos{θ} - 1)^{2}}{\sin{θ}} (-\sin{θ}) dθ$$ The \(\sin{θ}\) terms cancel out, leaving us with: $$-\int(\cos{θ}-1)^2 dθ$$ Now, we expand the square and integrate term by term: $$-\int(\cos^2{θ} - 2\cos{θ} +1) dθ = -\int \cos^2{θ} dθ +2\int \cos{θ} dθ -\int 1 dθ$$ We can evaluate these integrals separately: $$-\int \cos^2{θ} dθ = -\frac{1}{2} \int (1 + \cos{2θ}) dθ = -\frac{1}{2} (θ+\frac{1}{2}\sin{2θ}) + C_1$$ $$2\int \cos{θ} dθ = 2\sin{θ} + C_2$$ $$-\int 1 dθ = -θ + C_3$$ Adding the results of these integrals back together, we get: $$-\frac{1}{2} (θ+\frac{1}{2}\sin{2θ}) + 2\sin{θ} -θ + C = -\frac{1}{2}θ-\frac{1}{4}\sin{2θ}+2\sin{θ}-θ+C$$

Convert back to the original variable

Now we need to express our result in terms of \(x\) instead of \(θ\). Using our substitution, we can find the value of \(θ\): $$x+1=\cos{θ} \Rightarrow θ = \arccos(x+1)$$ We can also use the right triangle: $$\sin{θ} = \sqrt{1-\cos^2{θ}} = \sqrt{1-(x+1)^2} = -\sqrt{-(x+1)^2+1} = \sqrt{1-x^2-2x}$$ Then, we can find the value of \(\sin{2θ}\): $$\sin{2θ} = 2\sin{θ}\cos{θ} = 2\sqrt{1-x^2-2x}(x+1)$$ Finally, we substitute these values back into our result: $$-\frac{1}{2}\arccos(x+1)-\frac{1}{4}(2\sqrt{1-x^2-2x}(x+1))+2\sqrt{1-x^2-2x}-\arccos(x+1)+C$$ So the final answer is: $$-\frac{3}{2}\arccos(x+1)-\frac{1}{2}(x+1)\sqrt{1-x^2-2x}+2\sqrt{1-x^2-2x}+C$$