Question

第一题求解答不会做

Answer 1

∫ (sint)^4·(cost)^2 dt
＝∫(sint)^2·1/4·(sin(2t))^2dt
＝1/8∫(1-cos(2t))(sin(2t))^2dt
＝1/8∫(sin(2t))^2dt-1/16∫(sin(2t))^2d(sin(2t))
＝1/16∫1-cos(4t)dt-1/48 (sin(2t))^3
＝1/16 t-1/64 sin(4t)-1/48 (sin(2t))^3+C

追问

答非所问。