# integration by substitution/trig substitution

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

• Your substitution $x=tan o$ into the integral seems incorrect . You did not substitute $dx= sec^2 o do$. – Chinmaya mishra Oct 8 '19 at 6:12
• thank you everyone for kind reply, it sure put me back on the track – pikarin-g Oct 8 '19 at 8:17

\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}
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.
• $x\,\mathrm{d}x$ 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. – robjohn Oct 8 '19 at 14:00
$$\int 2 x^3 \sqrt{x^2+1} dx = 2 (\tan^3 t)\sqrt{\tan ^2 t+1}(1+\tan ^2 t) dt$$
You have ignored the $$dx = (1+\tan ^2 t) dt$$ part in your integral.