0
$\begingroup$

I attempted to find the integral of $\sec x$ via u substitution. This is what I did: $$\int \sec x dx = \int \frac{1}{\cos x} \ dx$$ Let $u = \cos x$

$$du = -\sin x \ dx$$ $$-sinx \int \frac{1}{\cos x} -\sin x \ dx$$ $$-\sin x \int \frac{1}{u} du = -\sin x \ ln|\cos x| + c$$

So why is this wrong? (I know what $\int \sec x \ dx$ is, and I've seen its derivation, but I don't understand where my attempt went wrong.)

$\endgroup$
  • $\begingroup$ How do you pull $\sin x$ out of the integral? $\endgroup$ – Moya Feb 24 '15 at 2:43
  • $\begingroup$ Because I needed -sinx for du, so I put -sinx outside the integral to balance the -sinx inside the integral. $\endgroup$ – Kat Feb 24 '15 at 2:45
  • $\begingroup$ That's not allowed since $\sin x$ depends on $x$, which you're integrating against. Or $\sin x=\sqrt{1-\cos^2(x)}=\sqrt{1-u^2}$, so it depends on $u$ as well. $\endgroup$ – Moya Feb 24 '15 at 2:47
3
$\begingroup$

When you do integration, you can only pull constants out. For example, $\int 2\sin x\, dx = 2 \int \sin x\, dx$, but $\int \sin x\cos x\, dx \neq \sin x \int \cos x\, dx$.

$\endgroup$
  • $\begingroup$ why can i only pull constants out? $\endgroup$ – Kat Feb 24 '15 at 2:51
  • $\begingroup$ @Kat look at the exmample I gave. Since $\int \sin x\cos x\, dx = \frac{1}{4}\cos(2x) + C$ and $\sin x \int \cos x\, dx = \sin^2x + D\sin x$, where $C$ and $D$ are constants, the two integrals are not equal. $\endgroup$ – kobe Feb 24 '15 at 2:55
  • $\begingroup$ oh, I see. Thanks for explaining. $\endgroup$ – Kat Feb 24 '15 at 3:01
1
$\begingroup$

You’ve not carried out the substitution correctly. Since $u=\cos x$ and $du=-\sin x\,dx$, we must have

$$dx=-\frac{du}{\sin x}=-\frac{du}{\sqrt{1-u^2}}\;,$$

and therefore

$$\int\sec x\,dx=\int\frac{du}{u\sqrt{1-u^2}}\;.$$

Even if you were allowed to move a function of $x$ through the integral sign, what you have would not be right: you want a factor of $-\sin x$, so the compensating factor should be $-\dfrac1{\sin x}$, not $-\sin x$. However, this is not a constant, so you cannot move it through the integral sign.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.