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Q&A

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

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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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How does $\sec\theta d\theta = d\left(\sec\theta\right)$? (1 comment)

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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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Self-answer as yourself (2 comments)

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