I am trying to solve this integral: $$\int{\frac{x^5}{\sqrt{x^2+7}}}\,\text{d}x.$$ I tried using the transform $x=\sqrt{7}\sinh{t}$ which leads to $$\int{(\sinh{t}})^5\,\text{d}t=\int \left(\frac{\cosh{2t}-1}{2}\right)^2 \times\sinh{t}\,\text{d}t.$$ There I can find the integrals except $\int\cosh[2 t]^2\times \sinh[t]\,\text{d}t$. How can I find this?
$\begingroup$
$\endgroup$
3
-
2$\begingroup$ I suggest starting over, let $u^2=x^2+7$. The problem collapses. $\endgroup$– André NicolasCommented Feb 5, 2015 at 12:05
-
$\begingroup$ yes , the problem does collapse , i see the denomiinator and start thinking trigonometric substitution $\endgroup$– avz2611Commented Feb 5, 2015 at 12:09
-
1$\begingroup$ There are some excellent integration strategies given in the answers below. However, if you (for whatever reason) were to insist on your original substitution, you can compute $$\int \cosh(2t)^2 \sinh(t)\ \mathrm{d}t = \frac{1}{2}\cosh(t) - \frac{1}{12}\cosh(3t) + \frac{1}{20}\cosh(5t) + C$$ simply by expanding $\sinh$ and $\cosh$ in terms of their exponential definitions, i.e., $$\sinh(x) = \frac{1}{2}(e^x - e^{-x}) \qquad \cosh(x) = \frac{1}{2}(e^x + e^{-x})$$ $\endgroup$– David ZhangCommented Feb 5, 2015 at 13:39
Add a comment
|
5 Answers
$\begingroup$
$\endgroup$
$$\int\frac{x^5}{\sqrt{x^2+7}}~dx~=~\int\frac{\big(x^2\big)^2~x}{\sqrt{x^2+7}}~dx~=~\frac12\int\frac{\Big[\big(x^2+7\big)-7\Big]^2}{\sqrt{x^2+7}}~d\big(x^2+7\big)=\int\frac{\big(t-7\big)^2}{2\sqrt t}~dt$$
Now let $t=u^2$, and use partial fraction decomposition. $\Big($Alternately, let $x=\sqrt7\cdot\sinh y\Big)$.
$\begingroup$
$\endgroup$
Hint: $$ \int\text{sinh}^5t\:\text{d}t=\int(\text{sinh}^2t)^2 \text{sinh}\:t \:\text{d}t = \int(\text{cosh}^2 t-1)^2 \:\text{d}(\text{cosh}\:t). $$
$\begingroup$
$\endgroup$
$$\int{\frac{x^5}{\sqrt{x^2+7}}}dx$$
Let $u = x^2 + 7 \implies du = 2x\,dx$ and $x^2 = u-7$.
Then $$\begin{align} \int{\frac{x^5}{\sqrt{x^2+7}}}dx & = \frac 12\int\frac {(x^2)^2 \,(2x\,dx)}{(x^2 + 7)^{1/2}} \\ \\ & = \frac 12\int \frac{(u-7)^2 \,du}{u^{1/2}}\\ \\ & = \frac 12\int\frac{u^2 - 14u + 49}{u^{1/2}}\,du \\ \\ &= \frac 12 \int \left(u^{3/2} - u^{1/2} + 49 u^{-1/2}\right)\,du\\ \\ &= \frac 12\left( \frac 25u^{5/2} - \frac 23 u^{3/2} + 2u^{1/2}\right) +C\\ \\ &= \frac 15(x^2-7)^{5/2} - \frac 13(x^2 - 7)^{3/2} + (x^2-7)^{1/2} +C\\ \\ &= \frac 1{15} \sqrt{x^2-7}\left(3(x^2-7)^2 - 5(x^2 -7) + 15\right) + C\\ \\ \end{align}$$
Of course, the last line can be simplified further.
$\begingroup$
$\endgroup$
2
take x as $\sqrt{7}\tan t$ and then proceed and second substitution take $\sec t=a$ these both should be enough
-
$\begingroup$ after posting this I though of using this transform. I am trying it now.. $\endgroup$ Commented Feb 5, 2015 at 12:10
-
$\begingroup$ it works all fine , but andres method is better , you get a polynomial with maybe a log term after integration $\endgroup$– avz2611Commented Feb 5, 2015 at 12:11
$\begingroup$
$\endgroup$
The comment of André Nicolas has anticipated my answer.....
The substitution: $$ x^2+7=t $$ gives: $$ 2xdx=dt $$ so we have: $$ \int{\dfrac{x^5}{\sqrt{x^2+7}}} dx= \int{\dfrac{x^5}{2x\sqrt{t}}} dt= \dfrac{1}{2}\int{\dfrac{(t-7)^2}{\sqrt{t}}} dt $$ that can be easily solved as a sum of powers.
If you want use your substitution note that: $$ \sinh^5(t)=\sinh t (\cosh^2 t-1)^2 $$ and the substitution: $$ \cosh t = u $$ gives $$ \sinh t dt =du $$ and you have again a polynomial....
answered Feb 5, 2015 at 12:18