2

$\int^{\pi/3} _0 \sin x \ln (\sec x)\text{dx}$

So far I have:

$u=\ln(\sec x), v'=\sin x$

$u'=\cos x, v=-\cos x$

$[-\cos x \ln (\sec x)-\int-\cos^2 x dx]^{\pi/3}_0$

Is this right so far?

André Nicolas
  • 507,029
Jim
  • 1,210

1 Answers1

11

Hint: better yet, try substituting $u=\cos{x}$. Then,

$$\begin{align}I&=\int^{\pi/3} _0 \sin x \ln (\sec x)\text{dx}\\ &=-\int^{\pi/3} _0 \sin x \ln (\cos x)\text{dx}\\ &=\int^{1/2} _1 \ln (u)\text{du}\end{align}$$

David H
  • 29,921
  • 1
    Thanks, very helpful! I got $[u \ln u - u]^{1/2}_1

    (\frac{1}{2} \ln\frac{1}{2} - \frac{1}{2})-(\ln 1-1)

    \frac{1}{2} \ln \frac{1}{2}+ \frac{3}{2}$, which is not the answer in the back: $\frac{1}{2}(1-ln2)$

    Is my answer correct?

    edit: my lines won't separate so it'll be hard to read, sorry...

     – Jim Jun 06 '14 at 19:45
  • 1
    @Jim It looks like you forgot to distribute negative sign. $-\frac12-(-1)\neq\frac32$. – David H Jun 06 '14 at 19:52
  • 1
    Ah thanks, even so, I get $\frac{1}{2}(\ln \frac{1}{2}+1)$. Is this correct? – Jim Jun 06 '14 at 19:54
  • 1
    @Jim Think about the basic algebraic rules for dealing with logarithms: $\log\frac12=\log 1 - \log 2$. See it now? – David H Jun 06 '14 at 19:57
  • 1
    Ah, thank you! I always forget to use that log law... – Jim Jun 06 '14 at 19:58