\begin{align*} \int\frac{\mathrm{d}x}{\cos(x) - 3\sin(x)} \end{align*}
I can't pick this numbers for formula $\cos(x)\cos(y) - \sin(x)\sin(y)$.
May you help me? Maybe I can use different way, but this way is more simple
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2Does this answer your question? Find $ \int \frac{1}{2\sin(x)-3\cos(x)}dx$. – José Carlos Santos May 19 '22 at 22:23
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You can combine the two trigonometric functions like so. – Тyma Gaidash May 20 '22 at 01:14
2 Answers
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To choose numbers as you want, you need that the sum of their squares is $1$. You can achieve this by writing $$ \frac1{\cos x-3\sin x}=\frac1{\sqrt{10}}\,\frac1{\frac1{\sqrt{10}}\,\cos x -\frac3{\sqrt{10}}\,\sin x}. $$
Martin Argerami
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Remember that:
$$\int \frac{dx}{\cos(x)-3\sin(x)}=\int\frac{dx}{-\sqrt{1+(-3)^3}\sin\left(x+\cot^{-1}(-3)\right)}=-10^{-\frac12}\int \csc\big(x-\cot^{-1}(3)\big)dx$$
Since $\frac d{dx}\ln\left(\tan\left(\frac {x+a}2\right)\right)=\csc(x+a)$:
$$\int \frac{dx}{\cos(x)-3\sin(x)}=c-10^{-\frac12} \ln\left(\tan\left(\frac {x-\cot^{-1}(3)}2\right)\right) $$
Please correct me and give me feedback!
Тyma Gaidash
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