f(x)=x^3 ; prove\:\tan^2(x)-\sin^2(x)=\tan^2(x)\sin^2(x) \frac{d}{dx}(\frac{3x+9}{2-x}) (\sin^2(\theta))' \sin(120) \lim _{x\to 0}(x\ln (x)) \int e^x\cos (x)dx \int_{0}^{\pi}\sin(x)dx

Therefore, if $\frac{dy}{dx}=f(x)$, then we can write it as y = $\int f(x)dx=F(x)+C$ where: $\int f(x)dx$ will represent the complete class of integral. C is an arbitrary constant. x is the variable …

\int_{0}^{2\pi}\cos^2(\theta)d\theta fracciones\:parciales\:\int_{0}^{1} \frac{32}{x^{2}-64}dx sustitución\:\int\frac{e^{x}}{e^{x}+e^{-x}}dx,\:u=e^{x}

derivative\:of\:f(x)=3-4x^2,\:\:x=5 implicit\:derivative\:\frac{dy}{dx},\:(x-y)^2=x+y-1 \frac{\partial}{\partial y\partial x}(\sin (x^2y^2))

x^2: x^{\msquare} \log_{\msquare} \sqrt{\square} \nthroot[\msquare]{\square} \le \ge \frac{\msquare}{\msquare} \cdot \div: x^{\circ} \pi \left(\square\right)^{'} \frac{d}{dx} …

x^2: x^{\msquare} \log_{\msquare} \sqrt{\square} \nthroot[\msquare]{\square} \le \ge \frac{\msquare}{\msquare} \cdot \div: x^{\circ} \pi \left(\square\right)^{'} \frac{d}{dx} …

\cos (x)-\sin (x)=0 \sin (4\theta)-\frac{\sqrt{3}}{2}=0,\:\forall 0\le\theta<2\pi ; 2\sin ^2(x)+3=7\sin (x),\:x\in[0,\:2\pi ] 3\tan ^3(A)-\tan (A)=0,\:A\in \:[0,\:360] 2\cos ^2(x)-\sqrt{3}\cos (x)=0,\:0^{\circ …

f(x)=x^3 ; provar\:\tan^2(x)-\sin^2(x)=\tan^2(x)\sin^2(x) \frac{d}{dx}(\frac{3x+9}{2-x}) (\sin^2(\theta))' \sin(120) \lim _{x\to 0}(x\ln (x)) \int e^x\cos (x)dx \int_{0}^{\pi}\sin(x)dx

Take the constant out \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx

x^2: x^{\msquare} \log_{\msquare} \sqrt{\square} \nthroot[\msquare]{\square} \le \ge \frac{\msquare}{\msquare} \cdot \div: x^{\circ} \pi \left(\square\right)^{'} \frac{d}{dx} …