Comments on Solve $\int_0^{\dfrac{\pi}{6}} \sec^3 \theta \mathrm d\theta$
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Solve $\int_0^{\dfrac{\pi}{6}} \sec^3 \theta \mathrm d\theta$
Evaluate $$\int_0^{\dfrac{\pi}{6}} \sec^3 \theta \mathrm d\theta$$
I was trying to solve it following way.
$$\int_0^{\dfrac{\pi}{6}} \sec^2\theta \sec\theta \mathrm d\theta$$ $$\int_0^{\dfrac{\pi}{6}}\sec^2\theta \mathrm d(\sec\theta)$$ $$[\tan\theta]_0^\dfrac{\pi}{6}$$ $$\tan\frac{\pi}{6}$$ $$\frac{1}{\sqrt{3}}$$
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.
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Is my answer correct?
My answer isn't correct. Cause, differentiation of $\sec x=\sec x\tan x$. I had differentiate inside integration.
$$\int_0^{\dfrac{\pi}{6}} \sec^3 \theta \mathrm d\theta$$ $$\int_0^{\dfrac{\pi}{6}} (1-\tan^2 \theta) \frac{d}{d \theta} (\sec \theta \tan \theta) \mathrm d\theta$$
That's the correct one. But, you got it wrong. I have differentiate.
my answer always doesn't match with their. So, is it OK to find new/another answer of Integral?
Saying to Indefinite Integral, if you integrate an equation then, I may find lots of answer. But, if I (you) put specific value instead of $\theta$ or, $x$. Than, you will get same value if your answer isn't wrong.
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