I want to solve the integral:$$\int x^2\sqrt{x^2+1}\;dx.$$ I did $x = \tan t$, then it is equal to:$$\int\frac{\tan^2 t}{\cos^3 t}\;dx.$$ Or:$$\int\frac{\sin^2 t}{\cos^5 t}\;dx.$$ I stuck there. Any help will be much appreciated...
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After the substitution shouldn't it be the following: $$\int \frac{ \tan^2(t)}{\cos^3(t)} dt$$ – Cahn Jan 01 '17 at 23:02
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Note that: $(\tan t)'=\frac{1}{\cos^2 x}$ – Olimjon Jan 01 '17 at 23:06
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1I know but $\int x^2 \sqrt{x^2+1} dx = \int \tan^2(t) \sqrt{\tan^2(t)+1} \sec^2(t) dt=\int \tan^2(t) \sec(t) \sec^2(t) dt$ – Cahn Jan 01 '17 at 23:09
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Yes, you are right. I made a mistake... – Olimjon Jan 01 '17 at 23:13
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Going the hyperbolic instead of the trigonometric route, we can let $x:=\sinh t$, then $dx=\cosh tdt$.
\begin{align} \int x^2\sqrt{x^2+1}dx &= \int\sinh^2t\sqrt{\sinh^2t+1}\cosh tdt\\ &=\int\sinh^2t\cosh^2tdt\\ &=\int\left(\sinh t\cosh t\right)^2dt\\ &=\frac{1}{4}\int\sinh^2(2t)dt\\ &=\frac{1}{8}\int\left(\cosh(4t)-1\right)dt\\ &=\frac{1}{32}\left(\sinh(4t)-4t\right) + C \end{align}
Since $t=\text{arsinh }x$ by inversion of our initial substitution, we have
\begin{align} \int x^2\sqrt{x^2+1}dx &= \frac{1}{32}\left[\sinh(4\text{arsinh }x)-4\text{arsinh }x\right]+C\\ &=\frac{1}{8}\left((x+2x^3)\sqrt{x^2+1}-\text{arsinh }x\right) + C \end{align}
If you want, you can write out $\text{arsinh }x$ using logarithms.
Tom
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An Euler substitution $$t=\sqrt{x^2+1}-x$$ will reduce the integral to a rational function.
Rene Schipperus
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You have to be a little hyperbolic savy to do this method but $x=\sinh u\to \text{d}x=\cosh u\text{d}u$ so the integral is now $\int \sinh^2 u\cosh^2 u\text{d}u=\int\frac{\sinh^2 2u}{4}\text{d}u=\int\frac{\cosh 4u-1}{8}\text{d}u=\frac{\frac{\sinh 4u}{4}-u}{8}=\frac{\sinh u \cosh u\cosh 2u-u}{8}=\frac{x\sqrt{1+x^2}\left ( 1+2x^2\right )-\sinh^{-1} x}{8}+C=\frac{x\sqrt{1+x^2}\left ( 1+2x^2\right )-\ln \left (x+\sqrt{1+x^2}\right )}{8}+C$
Teh Rod
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I liked your method, but I accepted another, as it was more easier to comprehend (for me). – Olimjon Jan 01 '17 at 23:25
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Yeah I honestly don't blame you haha, I just find using hyperbolics in integration extremely interesting – Teh Rod Jan 01 '17 at 23:25
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Try the following substitution $$x=\frac{e^t-e^{-t}}{2}$$ and you'll get an easy integral.
Michael Rozenberg
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$$\int\frac{\sin^2 t}{\cos^7 t}\;dt=\int\frac{\sin^2 t \cos t}{(1-\sin^2 t)^4}\;dt= \int \frac{u^2}{(1-u^2)^4} du \\ = \int \frac{u^2}{(1-u)^4(1+u)^4} du$$ and partial fractions.
An alternate way is to observe that $$\int\frac{\tan^2 t}{\cos^5 t}\;dt= \int \tan^2 t \sec^5 t dt$$ and use the reduction formula .
N. S.
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