Evaluation of $\int x^8\sqrt{x^2+1}\ dx$

Question

Evaluation of $\displaystyle \int x^8\sqrt{x^2+1}\ dx$

$\bf{My\: Try::}$ Put $x=\tan \theta\;,$ Then $dx = \sec^2 \theta d \theta$

So $$I = \int \sec ^{3}\theta \cdot \tan^{8}\theta d \theta=\int \frac{\sin^{8}\theta}{\cos^{5}\theta }d\theta$$

Now How can I solve it after that, Help Required, Thanks

Answer 1

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\begin{align} \color{#f00}{\int\root{x^{2} + 1}x^{8}\,\dd x} & = {1 \over 9}\int\root{x^{2} + 1}\,\dd\pars{x^{9}} \\ & = {1 \over 9}\,x^{9}\root{x^{2} + 1} - {1 \over 9}\int{x^{10} \over \root{x^{2} + 1}}\,\dd x \end{align}

With the sub$\ldots\quad$ $\ds{t \equiv \root{x^{2} + 1} - x\quad\imp\quad x = {1 - t^{2} \over 2t}}$: \begin{align} & \color{#f00}{\int\root{x^{2} + 1}x^{8}\,\dd x} = {1 \over 9}\,x^{9}\root{x^{2} + 1} - {1 \over 9216}\int{\pars{1 - t^{2}}^{10} \over t^{11}}\,\dd t \\[3mm] & = {1 \over 9}\,x^{9}\root{x^{2} + 1} - {1 \over 9216}\sum_{n = 0 \atop {\vphantom{\large A}n \not= 5}}^{10}{10 \choose n}\pars{-1}^{n} \int t^{2n - 11}\,\dd t + {7 \over 256}\ln\pars{t} \\[3mm] & = {1 \over 9}\,x^{9}\root{x^{2} + 1} - {1 \over18432}\sum_{n = 0 \atop {\vphantom{\large A}n \not= 5}}^{10} {10 \choose n}{\pars{-1}^{n} \over n - 5}\,t^{2n - 10} + {7 \over 256}\ln\pars{t} \\[3mm] & = \color{#f00}{{1 \over 9}\,x^{9}\root{x^{2} + 1} - {1 \over18432}\sum_{n = 0 \atop {\vphantom{\large A}n \not= 5}}^{10} {10 \choose n}{\pars{-1}^{n} \over n - 5}\,\pars{\root{x^{2} + 1} - x}^{2n - 10}} \\[3mm] & \color{#f00}{\phantom{==}+ {7 \over 256}\ln\pars{\root{x^{2} + 1} - x}} \end{align}

Answer 2

Hint: Using the identity $\tan^2\theta=\sec^2\theta-1$ you can rewrite your integral as $$\int{\sec^3(\theta)(\sec^2{(\theta)}-1)^4}\ d\theta$$

Answer 3

You can use Euler substitution: $\sqrt{x^2+1}=-x+t$. Then you get the integral:

$\int \sqrt{x^2+1}x^8dx = \int t*(\dfrac{t^2-1}{2t})^8*\dfrac{t^2+1}{2t^2}dt$

Answer 4

Expanding my comment:

setting $x=\sinh t$, $dx=\cosh t\ dt$, the integral becomes $$\int\sqrt{1+\sinh^2 t}\sinh^8 t\cosh t\ dt=\int\sinh^8t\cosh^2t\overset{P.I.}{=}\frac19\sinh^9t\cosh t-\int \sinh^{10}t\ dt$$ To evaluate this last integral you can use the definition $$\sinh t=\frac{e^t-e^{-t}}2$$ so that $$\sinh^{10} t=\frac{1}{2^{10}}\sum_{k=1}^{10}{10 \choose k}(-1)^ke^{(10-k)t}e^{-kt}=\frac{1}{2^{10}}\sum_{k=1}^{10}{10 \choose k}(-1)^ke^{(10-2k)t}$$

but I leave this up to you

Remark: of course you can substitute $\cosh^2 t$ in the second integral too, so that you then evaluate two $\sin^m t$ integrals. As done in the answer below

Answer 5

HINT:

Integrating by parts,

for $m+1\ne0,$

$$I_m=\int x^m\sqrt{x^2+1}\ dx=\sqrt{x^2+1}\int x^m\ dx-\int\left(\dfrac{d\sqrt{x^2+1}}{dx}\int x^m\ dx\right)dx$$

$$(m+1)I_m=x^{m+1}\sqrt{x^2+1}-\int\dfrac{x^{m+2}}{\sqrt{x^2+1}}dx$$

$$=x^{m+1}\sqrt{x^2+1}-\int\dfrac{x^m(1+x^2)-x^m}{\sqrt{x^2+1}}dx$$

$$(m+2)I_m=x^{m+1}\sqrt{x^2+1}+I_{m-2}-\int\dfrac{x^{m-2}}{\sqrt{x^2+1}}dx$$

The last part can be reduced further as $x^{m-2}=x^{m-4}(1+x^2)-x^{m-4}$

Answer 6

Let $x=\sinh t.$ Then $\sqrt {1+x^2}=\cosh t.$ And the integral is $$\int \cosh^2 t \sinh^8 t \;dt=\int (1+\sinh^2 t)\sinh^8 t dt=\int (\sinh^8 t+\sinh^{10}t) \;dt.$$ To evaluate $\int \sinh^{2 n}t \;dt $: For brevity let $S=\sinh t$ and $C=\cosh t,$ and let $\int \sinh^{2 n}t \;dt =V(2 n).$

Then $$V(2 n)=\int S^{2 n-1}\;dC=$$ $$=S^{2 n-1}C-\int C\;d(S^{2 n-1})=$$ $$=S^{2 n-1}C-\int (2 n-1)C^2 S^{2 n-2}\;dt=$$ $$=S^{2 n-1}C-\int (2 n-1)(1+S^2)S^{2 n-2}\;dt=$$ $$=S^{2 n-1}C-(2 n-1)(V(2 n-2)+V(2 n)).$$ This implies $$2 n V(2 n)=S^{2 n-1}C-(2 n-1)V(2 n-2).$$

Answer 7

By substituting $x=\sinh\theta$

$$ I = \int \cosh(\theta)^2 \sinh(\theta)^8\,d\theta = \int\sinh(\theta)^8\,d\theta+\int\sinh(\theta)^{10}\,d\theta$$ and the last integrals are straightforward to compute through De Moivre's formula, since

$$ J(n) = \int \sinh^{2n}(\theta)\,d\theta = \frac{1}{4^n}\sum_{k=0}^{2n}\binom{2n}{k}\int (-1)^k e^{-k\theta}e^{(2n-k)\theta}\,d\theta $$ gives: $$ J(n) = \frac{(-1)^n \theta}{4^n}\binom{2n}{n}+\frac{2}{4^n}\sum_{k=0}^{n-1}\binom{2n}{k}\frac{(-1)^k \cosh((2n-2k)\theta)}{2n-2k}.$$