A question regarding definite integrals

Asked May 01 '19 at 20:21

Active May 01 '19 at 20:24

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Why when I evaluate this integral:

$$ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{1}{\sin ^{2} x \cos ^{2} x} \mathrm{d} x $$

using my calculator in Radians I get an answer, whereas when I calculate it in Degrees like this:

$$ \int_{45}^{60} \frac{1}{\sin ^{2} x \cos ^{2} x} \mathrm{d} x $$

I get a different answer?

Aren't we just finding the same difference $F(b)-F(a)$?

asked May 01 '19 at 20:21

Positron12

You can't use degrees. If you use degrees, it is no longer true (for example) that $(\sin x)' = \cos x$. You would have to introduce a substitution that involved a factor of $180/\pi$. – MPW May 01 '19 at 20:24
Why isn't $$ (\sin x)^{\prime}=\cos x $$ in degrees? – Positron12 May 01 '19 at 20:27
See this question – MPW May 01 '19 at 20:33
Yeah, I got it, thank you . – Positron12 May 01 '19 at 20:35

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