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$$\int x(x^2+2)^4\,dx $$

When we do this integration with u substitution we get $$\frac{(x^2+2)^5}{10}$$ as $u=x^2+2$

$du=2x\,dx$ $$\therefore \int (u+2)^4\,du = \frac{(x^2+2)^5}{10} + C$$

Although when we expand the fraction and then integrate the answer we get is different:

$x(x^2+2)^4=x^9+8x^7+24x^5+32x^3+16x$ $$\int x^9+8x^7+24x^5+32x^3+16x \,dx$$

we get

$$\frac {x^{10}}{10} +x^8+4x^6+8x^4+8x^2 + C$$

For a better idea of the questions, let's say the questions asks us to find the value of C when y(0)=1

Now,

$x=0$

$$\frac {0^{10}}{10} + 0^8 + 4(0)^6 + 8(0)^4 + 8(0)^2 + C = 1$$ $$\therefore C= 1$$ AND $$\frac {(0+2)^5}{10} + C= 1$$ $$\therefore \frac {32}{10} + C = 1$$ $$\therefore C = 1 - 3.2 = -2.2$$

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    $\begingroup$ Don't forget the arbitrary constant of integration! $\endgroup$ – Lord Shark the Unknown Jun 12 '18 at 4:41
  • $\begingroup$ ^^^^^ What @LordSharktheUnknown said! The indefinite integral is the class of all functions who are antiderivatives of the integrand. These antiderivatives only differ by a constant. $\endgroup$ – N8tron Jun 12 '18 at 4:48
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    $\begingroup$ $\frac {(x^2+2)^5}{10} = \frac {x^{10}}{10} + x^8 + 4x^6+8x^4+8x^2 + \frac {32}{10}.$ The only difference is the constant of integration. $\endgroup$ – Doug M Jun 12 '18 at 4:52
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    $\begingroup$ Please fix the typos on $(x+2)$. $\endgroup$ – Yves Daoust Jun 12 '18 at 5:16
  • $\begingroup$ There are lots and lots of similar questions already, just search: google.com/… $\endgroup$ – Hans Lundmark Jun 12 '18 at 6:46
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Like mentioned in the comments this is all fixed if you remember your constant of integration.

$$\int x(x^2+2)^4\ dx= \frac{(x^2+2)^5}{10}+C$$

Note if you expand

$$ \begin{split} \frac{(x^2+2)^5}{10}&=\frac{1}{10}\left(x^{10}+5x^8(2)+10x^6(2^2)+10x^4(2^3)+5x^2(2^4)+2^5\right)\\ &=\frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2+\frac{32}{10} \end{split} $$

Notice the relation to your other way of computing the integral

$$ \int x(x^2+2)^4\ dx = \frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2 +C $$

So lets call $F(x)=\frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2$ and $G(x)=\frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2+\frac{32}{10}$ then $F(x)-G(x)=-\frac{32}{10}$ a constant. All antiderivatives of a continuous function only differ by a constant.


Just for fun Let's see another one:

First lets use double angle for sine $$ \int \cos x\sin x\ dx=\frac{1}{2}\int\sin 2x\ dx=-\frac{1}{4}\cos 2x +C $$

Then substitutions $u=\sin x$

$$ \int \cos x\sin x\ dx=\int u\ du =\frac{u^2}{2}+C=\frac{\sin^2 x}{2}+C $$

Then substitutions $u=\cos x$

$$ \int \cos x\sin x\ dx=\int -u\ du =\frac{-u^2}{2}+C=\frac{-\cos^2 x}{2}+C $$

If you find the constant differences and combine them in the right way you get the half angle formulas:

$$ \sin^2 x=\frac{1-\cos 2x}{2},\quad \cos^2 x=\frac{1+\cos 2x}{2} $$

Note you can pretty quickly derive some funky trig identities in this way. For instance if you consider $\int \cos^3 x \sin^5 x\ dx$

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  • $\begingroup$ Although if we were asked to find the value of C at x=0, won't we have 2 different value of C??? $\endgroup$ – Agent Smith Jun 13 '18 at 0:47
  • $\begingroup$ No if you fix any two antiderivatives $F$ and $G$ for a suitable function (continuous is adequate but the real condition is Riemann integrable). There is a constant C such that the difference $F(x)-G(x)=C$ for all x. This can be rigourously proven. $\endgroup$ – N8tron Jun 13 '18 at 1:45
  • $\begingroup$ Check out this page (or numerous others) to see the proof math.stackexchange.com/questions/1862231/… $\endgroup$ – N8tron Jun 13 '18 at 1:49
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You can check an antiderivative by differentiating.

$$\left(\frac{(x^2+2)^5}{10}\right)'=x(x^2+2)^4=x^9+8x^7+24x^5+32x^3+16x$$

and

$$\left(\frac {x^{10}}{10} +x^8+4x^6+8x^4+8x^2\right)'=x^9+8x^7+24+32x^3+16x$$

and the two expressions are indeed equivalent.


Now the long explanation.

Consider the binomial $x^2+a$ raised to some power $n$ and multiplied by $2x$.

$$2x(x^2+a)^m$$

which integrates as

$$\frac{(x^2+a)^{m+1}}{m+1}.$$

By the binomial theorem, the terms in the development of this antiderivative are

$$\frac1{m+1}\binom{m+1}kx^{2(m+1-k)}a^k.$$

On the other hand, the development of the initial integrand gives terms

$$2\binom mkx^{2(m-k)+1}a^k,$$ and after integration

$$\frac1{m-k+1}\binom mkx^{2(m-k)+2}a^k.$$

It is easy to see that all terms coincide, because

$$\frac1{m+1}\frac{(m+1)!}{k!(m+1-k)!}=\frac1{m-k+1}\frac{m!}{k!(m-k)!}=\frac{(m-1)!}{k!(m-k+1)!}.$$

Anyway, the first development holds for $0\le k\le m+1$, giving a constant term $\dfrac{a^m}{m+1}$, but the second for $0\le k\le m$ only, giving no constant term. But this does not matter, as two antiderivatives can differ by a constant.

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  • $\begingroup$ But wouldn't the C be different for both the cases?? $\endgroup$ – Agent Smith Jun 13 '18 at 0:48
  • $\begingroup$ @AgentSmith: and ? $\endgroup$ – Yves Daoust Jun 13 '18 at 5:50
  • $\begingroup$ and that would change the answer wouldn't it?? if we were asked for the value of C? $\endgroup$ – Agent Smith Jun 13 '18 at 19:56
  • $\begingroup$ @AgentSmith: you'd answer: arbitrary. You should understand that the value of the constant does not matter. $\endgroup$ – Yves Daoust Jun 13 '18 at 20:05
  • $\begingroup$ oh okay thanks! Makes sense! $\endgroup$ – Agent Smith Jun 14 '18 at 18:59

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