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i've been asked to integrating $2 x^3 \sqrt{x^2+1}$ and i did this by trig subtitution $ x= \tan{o} $ it turn

$$ \int 2 \tan^3{ o} \sqrt{\tan^2 {o} +1} do$$ by identity of trig and some customize

$$ \int 2 (\sec^2 o - 1) \sec o \tan o do $$ merge things up

$$ \int 2 ( \sec^2 o \sec o \tan o - \sec o \tan o do) $$ here i change $\sec o$ in front part with $u$ and $\sec o \tan o $ with $ du $ and doing power rule

$$ 2(\frac{1} {3} \sec^3 o - \sec o) $$ in tangent expression

$$ 2(\frac{1} {3} (1 + \tan^2 o) \sqrt {1 + \tan^2 o} - \sqrt {1 + \tan^2 o)} $$ and brings x back

$$ 2(\frac{1} {3} (1 + x^2) \sqrt {1 + x^2} - \sqrt {1 + x^2})$$

but my answer seems wrong when i checked it , so please tell me where did i go wrong

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  • $\begingroup$ Your substitution $x=tan o$ into the integral seems incorrect . You did not substitute $dx= sec^2 o do$. $\endgroup$ – Chinmaya mishra Oct 8 '19 at 6:12
  • $\begingroup$ thank you everyone for kind reply, it sure put me back on the track $\endgroup$ – pikarin-g Oct 8 '19 at 8:17
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$$ \begin{align} \int2x^3\sqrt{1+x^2}\,\mathrm{d}x &=\int2\tan^3(\theta)\sec(\theta)\,\mathrm{d}\color{#C00}{\tan(\theta)}\\ &=\int2\tan^3(\theta)\sec^3(\theta)\,\mathrm{d}\theta\\ &=\int2\tan^2(\theta)\sec^2(\theta)\,\mathrm{d}\sec(\theta)\\ &=\int2\left(\sec^2(\theta)-1\right)\sec^2(\theta)\,\mathrm{d}\sec(\theta)\\ &=\frac25\sec^5(\theta)-\frac23\sec^3(\theta)+C\\ &=\frac25\left(1+x^2\right)^{5/2}-\frac23\left(1+x^2\right)^{3/2}+C \end{align} $$

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You can also put $x^2+1=t^2$ or $2xdx=2tdt$ , Now your integral $$\int2x^2\sqrt{x^2+1}xdx$$ $$= 2\int(t^2-1)\sqrt{t^2}dt$$ Or $$2\int(t^3-t)dt$$ I think it is more simpler this way.

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  • $\begingroup$ $x\,\mathrm{d}x$ seems to have become $\mathrm{d}t$ instead of $t\,\mathrm{d}t$. Otherwise, this is a simpler approach, but it doesn't really address the question. $\endgroup$ – robjohn Oct 8 '19 at 14:00
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$$ \int 2 x^3 \sqrt{x^2+1} dx = 2 (\tan^3 t)\sqrt{\tan ^2 t+1}(1+\tan ^2 t) dt $$

You have ignored the $dx = (1+\tan ^2 t) dt $ part in your integral.

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