Search results
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} \frac{\partial}{\partial x} \int \int_{\msquare}^{\msquare} \lim \sum \infty \theta (f\:\circ\:g) f(x)
6 gru 2021 · $$\sec^2(x)=\tan^2(x)+1$$ you can attempt the change of variable $u=\tan(x)$ or $x=\arctan(u)$. This turns the integral to $$\int\frac{u^2+1}{u^2+1}du.$$
Indefinite integral. Step-by-step solution. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals.
17 mar 2016 · Answer link. tanx+C Note that 1/cosx=secx. Thus, int1/cos^2xdx=intsec^2xdx. This is a common integral: intsec^2xdx=tanx+C.
Free Integral Calculator helps you solve definite and indefinite integration problems. Also double, triple and improper integrals. Answers, graphs, alternate forms.
While integrating $f$, the author needs the inequality $$ \left|\frac{1-e^{iz}}{z^2}\right|\leq\frac{2}{|z|^2}, $$ which is justified by the fact that $|e^{iz}|\leq 1$ for $z$ in the upper half-plane.
Series expansion of the integral at x=0. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music….
