# Use Trigonometric Substitution to evaluate the integral. \int (x^2 \sqrt{ 3+2x-x^2}) dx

## Question:

Use Trigonometric Substitution to evaluate the integral.

{eq}\int (x^2 \sqrt{ 3+2x-x^2}) dx {/eq}

## Trigonometric Substitution:

The following trigonometric substitutions are effective for the given radical expressions:

{eq}\displaystyle \sqrt{x^2-a^2} \quad \quad x= a\sec t, \quad \quad 0 \leq t < \frac{\pi}{2} \text{ or } \pi \leq t < \frac{3\pi}{2} {/eq}

In each case, the restriction on {eq}t {/eq} is imposed to ensure that the function that defines the substitution is one-to-one.

We first rewrite {eq}3 + 2x - x^2 {/eq} by completing the square as follows:

{eq}\begin{align*} 3 + 2x - x^2 &= -\left(x^2 - 2x + -3\right)\\[0.3cm] &= -\left(x^2 - 2x + 1^2 - 1^2 - 3\right)\\[0.3cm] &= -\left(\left(x - 1\right)^2 -4\right)\\[0.3cm] &= 4 - \left(x - 1\right)^2 \end{align*} \\ {/eq}

Now, let {eq}\displaystyle x-1 = 2\sin t {/eq}, where {eq}\displaystyle -\frac{\pi}{2}\leq t\leq \frac{\pi}{2} {/eq}. Then {eq}\displaystyle dx = 2\cos t \, dt {/eq} and {eq}\displaystyle t = \sin^{-1}\left(\frac{x-1}{2}\right) {/eq}. Hence,

{eq}\begin{align*} \int x^2 \sqrt{3 + 2x - x^2} \, dx &= \int x^2\sqrt{4 - (x-1)^2}\, dx\\[0.3cm] &= \int\left(2\sin t + 1\right)^2\sqrt{4 - 4\sin^2 t}\,2\cos t\, dt\\[0.3cm] &=2\int\left(4\sin^2t + 4\sin t + 1\right)\sqrt{4(1 - \sin^2t)}\cos t\, dt\\[0.3cm] &= 4\int\left(4\sin^2t + 4\sin t + 1\right)\sqrt{\cos^2t}\cos t\, dt\\[0.3cm] &= 4\int\left(4\sin^2t + 4\sin t + 1\right)\cos t\cos t\, dt & \left(\text{since } -\frac{\pi}{2}\leq t\leq \frac{\pi}{2}\right)\\[0.3cm] &= 4\int\left(4\sin^2t + 4\sin t + 1\right)\cos^2 t\, dt\\[0.3cm] &= 4\int\left(4\sin^2t\cos^2t + 4\sin t\cos^2 t + \cos^2 t\right)\, dt\\[0.3cm] &= 4\int\left(4(1 - \cos^2 t)\cos^2t + 4\sin t\cos^2 t + \cos^2 t\right)\, dt\\[0.3cm] &= 4\int\left(4\cos^2 t -4 \cos^4 t + 4\sin t\cos^2 t + \cos^2 t\right)\, dt\\[0.3cm] &= 4\int\left(5\cos^2 t -4 (\cos^2 t)^2 + 4\sin t\cos^2 t \right)\, dt\\[0.3cm] &= 4\int\left(\frac{5}{2}\left(\cos 2 t + 1\right) -4 \left(\frac{1}{2}\left(\cos 2 t + 1\right)\right)^2 + 4\sin t\cos^2 t \right)\, dt & \left(\cos^2t = \frac{1}{2}(\cos 2t + 1)\right)\\[0.3cm] &= 4\int\left(\frac{5}{2}\left(\cos 2 t + 1\right) - \cos^2 2 t - 2\cos 2t - 1+ 4\sin t\cos^2 t \right)\, dt\\[0.3cm] &= 4\int\left(\frac{1}{2}\cos 2 t + \frac{3}{2}- \cos^2 2 t+ 4\sin t\cos^2 t \right)\, dt\\[0.3cm] &= 4\int\left(\frac{1}{2}\cos 2 t + \frac{3}{2}- \frac{1}{2}(\cos 4 t + 1)+ 4\sin t\cos^2 t \right)\, dt & \left(\cos^22t = \frac{1}{2}(\cos 4t + 1)\right)\\[0.3cm] &= 4\int\left(\frac{1}{2}\cos 2 t + 1- \frac{1}{2}\cos 4 t + 4\sin t\cos^2 t \right)\, dt \\[0.3cm] &= 4\left(\frac{1}{4}\sin 2t + t -\frac{1}{8}\sin 4t - \frac{4}{3}\cos^3 t\right) + C\\[0.3cm] &= 4\left(\frac{1}{2}\sin t\cos t + t -\frac{1}{4}\sin 2t\cos 2t - \frac{4}{3}\cos^3 t\right) + C\\[0.3cm] &= 4\left(\frac{1}{2}\sin t\cos t + t -\frac{1}{2}\sin t\cos t(1- 2\sin^2 t) - \frac{4}{3}\cos^3 t\right) + C\\[0.3cm] & =2 \sin t \cos t + 4t - 2 \sin t \cos t + 4 \sin^3t\cos t - \frac{16}{3}\cos^3 t + C\\[0.3cm] &= 4t + 4 \sin^3t\cos t - \frac{16}{3}\cos^3 t + C\\[0.3cm] & = 4t + 4 \sin^3t\sqrt{1 - \sin^2t} - \frac{16}{3}\left(1 - \sin^2t\right)^{\frac{3}{2}} + C & (\sin^2t + \cos^2 t = 1)\\[0.3cm] &= 4\sin^{-1}\left(\frac{x-1}{2}\right) +4 \left(\frac{x-1}{2}\right)^3\sqrt{1 - \frac{(x-1)^2}{4}} -\frac{16}{3}\left(1 - \frac{(x-1)^2}{4} \right)^{\frac{3}{2}} + C\\[0.3cm] &= 4\sin^{-1}\left(\frac{x-1}{2}\right) +4 \left(\frac{x-1}{2}\right)^3\sqrt{\frac{1}{4}\left(4 - (x-1)^2\right)} -\frac{16}{3}\left(\frac{1}{4}\left(4 - (x-1)^2 \right)\right)^{\frac{3}{2}} + C\\[0.3cm] &= 4\sin^{-1}\left(\frac{x-1}{2}\right) +2 \left(\frac{x-1}{2}\right)^3\sqrt{4 - (x-1)^2} -\frac{2}{3}\left(4 - (x-1)^2\right)^{\frac{3}{2}} + C\\[0.3cm] &= 4\sin^{-1}\left(\frac{x-1}{2}\right) + \frac{(x-1)^3}{4}\sqrt{3 + 2x - x^2} -\frac{2}{3}\left(3 + 2x - x^2\right)^{\frac{3}{2}} + C\\ \end{align*} {/eq}