indefinite-integration

Question: $\int_{{}}^{{}}{x{{\cos }^{2}}}xdx=$

1) $\frac{{{x}^{4}}}{4}-\frac{1}{4}x\sin 2x-\frac{1}{8}\cos 2x+c$

2) $\frac{{{x}^{2}}}{4}+\frac{1}{4}x\sin 2x+\frac{1}{8}\cos 2x+c$

3) $\frac{{{x}^{4}}}{4}-\frac{1}{4}x\sin 2x+\frac{1}{8}\cos 2x+c$

4) $\frac{{{x}^{4}}}{4}+\frac{1}{4}x\sin 2x-\frac{1}{8}\cos 2x+c$

Solution:

Explanation: No Explanation

Integration by parts

Rate this question:

★ ★ ★ ★ ★

Average rating: (0 votes)