To compute improper integral $\int_3^{5}\frac{x^{2}\, dx}{\sqrt{x-3}{\sqrt{5-x}}}$

Question

I am given improper integral as

$$\int_3^{5}\frac{x^{2}}{\sqrt{x-3}{\sqrt{5-x}}}dx$$

DOUBT

I see that problem is at both the end points, so i need to split up the integral. But problem seems to me that when i split up integral one of terms in denominator which is ${5-x}$ becomes negative. So how do i split up the integral ? Kindly help

Thanks

Answer 1

I'd suggest that you do $u=x-4$ to get an integral from $-1$ to $1$. It will be $$ \int_{-1}^1\frac{(u+4)^2}{\sqrt{1-u^2}}\,du=\int_{-1}^1-\sqrt{1-u^2}+\frac{8u}{\sqrt{1-u^2}}+\frac{17}{\sqrt{1-u^2}}\,du. $$ The first term in the integrand gives $-\pi/2$, since it represents (minus) the area of a half circle. The second term gives nothing, since it is odd, and the interval is even. The last term gives $$ \bigl[17\arcsin u\bigr]_{-1}^1=17(\pi/2-(-\pi/2))=17\pi. $$ All in all, the integral equals $-\pi/2+17\pi=33\pi/2$.

Answer 2

Beta Function Approach

Substituting $x\mapsto2x+3$, $$ \begin{align} &\int_3^5\frac{x^2\,\mathrm{d}x}{\sqrt{(x-3)(5-x)}}\\ &=\int_0^1\frac{(2x+3)^2\,\mathrm{d}x}{\sqrt{x(1-x)}}\\ &=\int_0^1\frac{(3(1-x)+5x)^2\,\mathrm{d}x}{\sqrt{x(1-x)}}\\ &=9\int_0^1\frac{(1-x)^2\,\mathrm{d}x}{\sqrt{x(1-x)}} +30\int_0^1\frac{(1-x)x\,\mathrm{d}x}{\sqrt{x(1-x)}} +25\int_0^1\frac{x^2\,\mathrm{d}x}{\sqrt{x(1-x)}}\\ &=9\operatorname{B}\left(\frac52,\frac12\right) +30\operatorname{B}\left(\frac32,\frac32\right) +25\operatorname{B}\left(\frac12,\frac52\right)\\ &=9\cdot\frac38\pi +30\cdot\frac18\pi +25\cdot\frac38\pi\\ &=\frac{33}2\pi \end{align} $$ using the Beta Function.

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Trigonometric Subsitution

After the substitution $x\mapsto2x+3$, we can use $x\mapsto\sin^2(x)$ $$ \begin{align} &\int_3^5\frac{x^2\,\mathrm{d}x}{\sqrt{(x-3)(5-x)}}\\ &=9\int_0^1\frac{(1-x)^2\,\mathrm{d}x}{\sqrt{x(1-x)}} +30\int_0^1\frac{(1-x)x\,\mathrm{d}x}{\sqrt{x(1-x)}} +25\int_0^1\frac{x^2\,\mathrm{d}x}{\sqrt{x(1-x)}}\\ &=9\int_0^{\pi/2}2\cos^4(x)\,\mathrm{d}x +30\int_0^{\pi/2}2\sin^2(x)\cos^2(x)\,\mathrm{d}x +25\int_0^{\pi/2}2\sin^4(x)\,\mathrm{d}x\\ &=9\cdot\frac38\pi +30\cdot\frac18\pi +25\cdot\frac38\pi\\ &=\frac{33}2\pi \end{align} $$

Answer 3

Hint: $$(5-x)(x-3)=(-x^2-15+8x)=(1-(x-4)^2)$$ now let $x=sint$ and you will get rid of the square root. And be careful about the range, which changes when you changing your variable.

Answer 4

Notice, the following

$$x^2=A(x-3)(5-x)+B(8-2x)+C$$

On solving we get

$$x^2=-(x-3)(5-x)-4(8-2x)+17$$ Now, we have $$\int_{3}^{5}\frac{x^2dx}{\sqrt{(x-3)}\sqrt{(5-x)}}$$ $$=\int_{3}^{5}\frac{(-(x-3)(5-x)-4(8-2x)+17)dx}{\sqrt{(x-3)}\sqrt{(5-x)}}$$ $$=-\int_{3}^{5}\frac{(x-3)(5-x)dx}{\sqrt{(x-3)}\sqrt{(5-x)}}-4\int_{3}^{5}\frac{(8-2x)dx}{\sqrt{(x-3)}\sqrt{(5-x)}}+17\int_{3}^{5}\frac{dx}{\sqrt{(x-3)}\sqrt{(5-x)}}$$ $$=-\int_{3}^{5}\sqrt{(x-3)}\sqrt{(5-x)}-4\int_{3}^{5}\frac{(8-2x)dx}{\sqrt{8x-x^2-15}}+17\int_{3}^{5}\frac{dx}{\sqrt{1-(x-4)^2}}$$ $$=-\int_{3}^{5}\sqrt{1-(x-4)^2}-4\int_{3}^{5}\frac{(8-2x)dx}{\sqrt{8x-x^2-15}}+17\int_{3}^{5}\frac{dx}{\sqrt{1-(x-4)^2}}$$ $$=-\frac{1}{2}\left[(x-4)\sqrt{1-(x-4)^2}+\sin^{-1}\left(x-4\right)\right]_{3}^{5}-4\left[2\sqrt{1-(x-4)^2}\right]_{3}^{5}+17\left[\sin^{-1}(x-4)\right]_{3}^{5}$$ $$=-\frac{1}{2}(\pi)-8(0)+17(\pi)=\frac{33\pi}{2}$$