NettetSolve ∫ sin^5 (2t)cos^2 (2t)dt Microsoft Math Solver ∫ (sin(2t))5(cos(2t))2dt Evaluate 764(sin(t))14 − 32 (sin(t))12 + 5208(sin(t))10 − 24 (sin(t))8 + 316(sin(t))6 + С … Nettet8. sep. 2024 · cos 5 ( t) = cos ( t) ( 1 + cos ( 2 t) 2) 2 = 1 4 [ cos ( t) ( 1 + 2 cos ( 2 t) + cos 2 ( 2 t))] Then simplifying the integral: 1 4 [ ∫ cos ( t) ( sin ( t)) 1 3 + 2 ∫ cos ( 2 t) ( sin ( t)) 1 3 + ∫ cos 2 ( 2 t) ( sin ( t)) 1 3] d t Using u substitution for the first integral as u = sin ( t) and d u = cos ( t) d t

Selesaikan 6sin(2t)+6sin(t)=0 Microsoft Math Solver

Nettet11. apr. 2015 · f (x) = ∫ sin(2t + 1) cos2(2t +1) dt. Let's u = 2t +1. du = 2. Now we have : f (x) = 1 2 ∫sin(u) ⋅ cos−2(u)du. f (x) = − 1 2∫ − sin(u) ⋅ cos−2(u)du. F (x) = 1 2 [cos−1(u)] … Nettet使用包含逐步求解过程的免费数学求解器解算你的数学题。我们的数学求解器支持基础数学、算术、几何、三角函数和微积分 ... john green facebook middlesbrough

다음을 풀어보세요: cos^2t+sin^2t Microsoft Math Solver

Nettet2,433 solutions calculus Evaluate the definite integral. ∫_0^π/2 cos t / (1 + sin t) dt calculus Convert the point from cylindrical coordinates to rectangular coordinates. (2, -π, -4) calculus Evaluate the integral. ∫ 3t-2 / t+1 dt calculus calculus Find an … Nettet14. aug. 2015 · ∫ 2 t 8 + 5 cos 2 t d t In an attempt to apply integration by parts, we have u = 2 t ⇒ d u = 2 d t v = ∫ 8 + 5 cos 2 t d t = ∫ 13 − 5 sin 2 t d t = 13 ∫ 1 − 5 13 sin 2 t d t = 13 E ( t 5 13) + C So now we end up with 2 13 t E ( t 5 13) − 2 13 ∫ E ( t 5 13) d t Where E is the incomplete elliptic integral of the second kind. NettetAnvendelser Hypotenus Funksjoner Inverse funksjoner Referanse Identiteter Eksakte konstanter Trigonometriske tabeller Setninger Sinussetningen Cosinussetningen … interarms 38 spl