定积分∫(x^2arctanx+cos^5x)dx

定积分∫(x^2arctanx+cos^5x)dx
定积分∫(x^2arctanx+cos^5x)dx

定积分∫(x^2arctanx+cos^5x)dx
∫ x²arctanx dx + ∫ cos⁵x dx
= ∫ arctanx d(x³/3) + ∫ cos⁴x d(sinx)
= (1/3)x³arctanx - (1/3)∫ x³ d(arctanx) + ∫ (1 - sin²x)² d(sinx)
= (1/3)x³arctanx - (1/3)∫ x[(1 + x²) - 1]/(1 + x²) dx + ∫ (1 - 2sin²x + sin⁴x) d(sinx)
= (1/3)x³arctanx - (1/3)∫ [x - x/(1 + x²)] dx + sinx - (2/3)sin³x + (1/5)sin⁵x
= (1/3)x³arctanx - x²/6 + (1/6)ln(1 + x²) + sinx - (2/3)sin³x + (1/5)sin⁵x + C
Hope my answer will be helpful...

不定积分。。。题目写清楚一点