2024年12月16日 · Ex 3.3, 23 Prove that tan⁡4𝑥 = (4 tan⁡〖𝑥 (1−tan2𝑥)〗)/(1 − 6 tan2 𝑥+tan4 𝑥) Solving L.H.S. tan 4x We know that tan 2x = (2 𝑡𝑎𝑛⁡𝑥)/(1 − 𝑡𝑎𝑛2 𝑥) Replacing x with 2x tan (2 × 2x) = (2 𝑡𝑎𝑛⁡2𝑥)/(1 − 𝑡𝑎𝑛2 2𝑥) tan 4x = (2 𝑡𝑎𝑛⁡2𝑥)/(1 − 𝑡𝑎𝑛2 2𝑥 ...

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Find the derivatives of \frac{d}{dx}\int _2^{x^3}\sin \left(t^2\right)dt and \frac{d}{dx}\int _{\tan x}^{\sqrt{x+2}}\sqrt[3]{1+t^2}dt https://math.stackexchange.com/questions/2258131/find-the-derivatives-of-fracddx-int-2x3-sin-leftt2-rightdt-and

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2020年12月2日 · $$\tan (x) = \tan (4x)\xrightarrow{{n \in \mathbb{Z}}}4x = n\pi + x \to 3x = n\pi \to x = \frac{{n\pi }}{3}$$

2010年6月26日 · 另一种方法是利用公式：1+tan²x=sec²x将tan^4(x)化为： tan^4(x)=(sec²-1)²=sec^4(x)-2sec²x+1 上式后两项的积分易求。第一项的积分可写为,

Use the form atan(bx−c)+ d a tan (b x - c) + d to find the variables used to find the amplitude, period, phase shift, and vertical shift. Since the graph of the function tan t a n does not have a maximum or minimum value, there can be no value for the amplitude. Find the period of tan(4x) tan (4 x). Tap for more steps...

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$$=\dfrac{2(\dfrac{2\tan x}{1-\tan^2x})}{1-(\dfrac{2\tan x}{1-\tan^2x})^2}$$ $$=\dfrac{4\tan x (1-\tan^2x)}{(1-\tan^2x)^2-4\tan^2x}$$ $$=\dfrac{4\tan x(1-\tan^2x)}{1-6\tan^2x+\tan^4x}$$

2024年12月16日 · Ex 7.3, 16 ∫1 〖tan^4 𝑥〗 𝑑𝑥 ∫1 〖tan^4 𝑥〗 𝑑𝑥=∫1 〖tan^2 𝑥 .tan^2 𝑥〗 𝑑𝑥 =∫1 〖(sec^2⁡𝑥− 1) tan^2⁡𝑥 ...