Solve $\int_0^{\pi/6} \sec^3\theta\, \mathrm d\theta$

Question

Evaluate $$\int_0^{\frac{\pi}{6}} \sec^3 \theta \mathrm d\theta$$

I was trying to solve it following way.

$$\int_0^{\frac{\pi}{6}} \sec^2\theta \sec\theta \mathrm d\theta$$ $$\int_0^{\frac{\pi}{6}}\sec^2\theta \mathrm d(\sec\theta)$$ $$[\tan\theta]_0^\dfrac{\pi}{6}$$ $$\tan\frac{\pi}{6}$$ $$\frac{1}{\sqrt{3}}$$

I had found the value. But, my book had solved it another way. They took

$$\tan\theta=z$$ Then, they solved it. They had got $\frac{1}{3}+\frac{1}{2}\ln\sqrt{3}$. My answer is approximately close to their. Is my answer correct? While doing Indefinite integral I saw that I could solve problem my own way. But, my answer always doesn't match with their. So, is it OK to find new/another answer of Integral? In algebraic expression,"no matter what I do the answer always matches". But, I got confused with Integration.

I was checking whole process of my book. I saw something wasn't looking perfect.

$$\int_0^{\frac{\pi}{6}} \sqrt{1+\tan^2\theta}\sec^2\theta \mathrm{d\theta}$$

But, $\sqrt{1+\tan^2\theta}\sec^2\theta \mathrm{d\theta}\neq sec^3\theta$

Isn't it wrong?

@KaviRamaMurthy Suddenly, the book is like,"Am I joke to you?" — , Jul 27 '21 at 10:04
@KaviRamaMurthy No, it's not. The correct answer is $\frac13+\frac{\log 3}4\left(=\frac13+\frac{\log\left(\sqrt3\right)}2\right)$. — José Carlos Santos, Jul 27 '21 at 10:07
@Istiak You seem to state that $d(\sec \theta) = \sec \theta d\theta$ which is not correct. — mathcounterexamples.net, Jul 27 '21 at 10:09
@Istiak Your first step somehow changed $\sec\theta \ d\theta \to d(\sec\theta) \to d\theta$. — peterwhy, Jul 27 '21 at 10:09
@mathcounterexamples.net $\sec\theta \mathrm d \theta=\frac{d}{d\theta} (\sec\theta)d\theta=d(\sec\theta)$ — , Jul 27 '21 at 10:10
Sorry .... but it is wrong, alas. $\frac d{dt}\sec(t)=\sec(t)\tan(t)$ — Claude Leibovici, Jul 27 '21 at 10:12
@Istiak Where does $\sec\theta \mathrm d \theta=\frac{d}{d\theta} (\sec\theta)d\theta$ comes from??? — mathcounterexamples.net, Jul 27 '21 at 10:13
@mathcounterexamples.net I was differentiating. But, I forgot to differentiate $\sec$ — , Jul 27 '21 at 10:14
@Istiak About your edit, $\sqrt{1+\tan^2\theta}\sec^2\theta = \sec^3\theta$ because $1+\tan^2\theta = \sec^2\theta$, and $\sec\theta\ge0$ in the range. — peterwhy, Jul 27 '21 at 10:29

score 1 · Answer 1 · answered Jul 29 '21 at 01:38

1

No, it is not wrong because of the fact that $1+\tan^2{x}=\sec^2{x}$ and $\sqrt{1+\tan^2{x}}$ is just $\sec{x}$. Please actually try the problem and look for any potential mistakes before resorting to Math Stack Exchange. :)

answered Jul 29 '21 at 01:38

Arjun Vyavaharkar

497

Solve $\int_0^{\pi/6} \sec^3\theta\, \mathrm d\theta$

1 Answers1