将函数\(f(x)=\sin^4 x\)展开成Fourier级数为 ____ .
A: \(f(x) = \frac{3}{8}-\frac{1}{2}\cos 2x +\frac{1}{8}cos 4x\)
B: \(f(x) = \frac{1}{4}-\frac{1}{2}\cos x +\frac{3}{8}cos 4x\)
C: \(f(x) = \frac{1}{4}-\frac{1}{2}\sin 2x -\frac{3}{8}cos 4x\)
D: \(f(x) = \frac{3}{8}-\frac{1}{2}\sin x -\frac{1}{8}cos 4x\)
A: \(f(x) = \frac{3}{8}-\frac{1}{2}\cos 2x +\frac{1}{8}cos 4x\)
B: \(f(x) = \frac{1}{4}-\frac{1}{2}\cos x +\frac{3}{8}cos 4x\)
C: \(f(x) = \frac{1}{4}-\frac{1}{2}\sin 2x -\frac{3}{8}cos 4x\)
D: \(f(x) = \frac{3}{8}-\frac{1}{2}\sin x -\frac{1}{8}cos 4x\)
举一反三
- 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 \sin^3 x \cos x dx = $ A: $\frac{\sin^4 x}{4} +C$ B: ${\sin^4 x} +C$ C: $\frac{\cos^4 x}{4} +C$ D: $\frac{\cos^4 x}{4} +C$
- 求下列不定积分.[tex=7.286x2.643]28VI4S//fW038PiMAbBHktfj3FfJYocy4+TgcP5gH+6DCjcL5MVe5w4GLCJx2oaC[/tex].腺 由于 $\sin ^{4} x+\cos ^{4} x=\left(\cos ^{2} x-\sin ^{2} x\right)^{2}+2 \sin ^{2} x \cos ^{2} x$$=\cos ^{2} 2 x+\frac{1}{2} \sin ^{2} 2 x$原式 $=\int \frac{\mathrm{d} x}{\cos ^{2} 2 x+\frac{1}{2} \sin ^{2} 2 x}$
- \(\int { { {\sin }^{2}}x { { \cos }^{5}}xdx}\)=( ) A: \(\frac{1}{3} { { \sin }^{3}}x-\frac{2}{5} { { \sin }^{5}}x+\frac{1}{7} { { \sin }^{7}}x+C\) B: \(\frac{2}{3} { { \sin }^{3}}x-\frac{1}{5} { { \sin }^{5}}x-\frac{1}{7} { { \sin }^{7}}x+C\) C: \(\frac{1}{3} { { \cos }^{3}}x-\frac{2}{5} { { \cos }^{5}}x+\frac{1}{7} { { \cos }^{7}}x+C\) D: \(\frac{2}{3} { { \cos }^{3}}x-\frac{1}{5} { { \cos }^{5}}x-\frac{1}{7} { { \cos }^{7}}x+C\)
- 积分\(\int_0^1 (x\sin\frac{1}{x^2} - \frac{1}{x}\cos\frac{1}{x^2})dx\) (不计算积分, 由判别法直接判断)