积分$\int_0^1 x \arctan xdx=$（）。
A: $\frac{\pi}{4}+\frac{1}{2}$
B: $\frac{\pi}{4}$
C: $\frac{\pi}{4}-\frac{1}{2}$
D: $\frac{1}{2}$
A: $\frac{\pi}{4}+\frac{1}{2}$
B: $\frac{\pi}{4}$
C: $\frac{\pi}{4}-\frac{1}{2}$
D: $\frac{1}{2}$
举一反三
- Solve $ \frac{1}{\pi}\int_0^{\frac{\pi}{2}}\sin^4{x}dx=$ :<br/>______
- Solve $\int_{-\frac{1}{2}}^1{1-x^2}dx=$? A: $\frac{\pi}{3}+\frac{\sqrt{3}}{8}$. B: $\frac{\pi}{2}$. C: $\frac{\pi}{6}+\frac{\sqrt{3}}{4}$. D: $\frac{\pi}{4}$.
- For the integral $\int_0^{+\infty}\frac{dx}{(x^2+p^2)(x^2+q^2)}$, which of the following statements are CORRECT? A: $\frac{1}{q^2-p^2}[\frac{1}{p}-\frac{1}{q}]\frac{\pi}{2},p>0 \ q>0;$ B: $\frac{1}{q^2-p^2}[\frac{1}{q}+\frac{1}{p}]\frac{\pi}{2}, -p>0 \ -q>0;$ C: $\frac{1}{q^2-p^2}[\frac{1}{p}-\frac{1}{q}]\frac{\pi}{2}, p>0 \ -q>0;$ D: $\frac{1}{p^2-q^2}[\frac{1}{q}+\frac{1}{p}]\frac{\pi}{2}, -p>0 \ q>0.$
- 函数\(f(x) = x^2,\; x \in [-\pi,\pi]\)的Fourier级数为 A: \(\frac{\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \sin nx ,\; x \in [-\pi,\pi]\) B: \(\frac{\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx ,\; x \in [-\pi,\pi]\) C: \(\frac{2\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \sin nx ,\; x \in [-\pi,\pi]\) D: \(\frac{2\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx ,\; x \in [-\pi,\pi]\)
- $\int_{0}^{\frac{\text{ }\!\!\pi\!\!\text{ }}{4}}{[\cos (2t)\mathbf{i}+\sin (2t)\mathbf{j}+t\sin t\mathbf{k}]}\operatorname{dt}=$( ) A: $(\frac{1}{2},\frac{1}{2},\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$ B: $(1,\frac{1}{2},\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$ C: $(\frac{1}{2},1,\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$ D: $(1,1,\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$