Integral of sin 6x+cos 6x/sin 2x cos 2x
Nettet16. nov. 2016 · How can we simplify $\cos^6x+\sin^6x$ to $1−3\sin^2x\cos^2x$?. One reasonable approach seems to be using $\left(\cos^2x+\sin^2x\right)^3=1$, since it contains the terms $\cos^6x$ and $\sin^6x$.Another possibility would be replacing all occurrences of $\sin^2x$ by $1-\cos^2x$ on both sides and then comparing the … Nettet3. feb. 2024 · hint: $\sin^2 x + \cos^2 x = 1$; if you square that, you get $$ \sin^4 x + 2\sin^2 x \cos^2 x + \cos^4 x = 1, $$ which you can turn into $$ \sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x $$ and that lets you get rid of the 4th powers in the left hand side.. Can you run with that idea?
Integral of sin 6x+cos 6x/sin 2x cos 2x
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Nettet20. sep. 2024 · asked Jun 25, 2024 in Indefinite Integral by Siwani01 (50.7k points) integrals; class-12; 0 votes. 1 answer. Evaluate: ∫dx/√(x^2 - 6x + 10) ... Evaluate : … Nettet12. mar. 2016 · We should try to use substitution by setting u = cosx, so du = − sinxdx. This gives us the integral: ∫ sinx cos2x dx = − ∫ −sinx cos2x dx = −∫ 1 u2 du = − …
NettetTrigonometry sin(4x)−sin(6x)+ cos(2x)+cos(4x)−cos(6x) = Videos 05:38 Explicación de la propiedad distributiva (artículo) Khan Academy khanacademy.org Introducción a las derivadas parciales (artículo) Khan Academy khanacademy.org 08:30 Simplificar expresiones con raíz cuadrada Khan Academy Integrales triples (artículo) Khan …
NettetSolution Verified by Toppr Correct option is C) To find ∫cos 8xsin 6x⋅dx=∫tan 6x⋅sec 2x⋅dx, substitute tanx=t sec 2xdx=dt =∫t 6dt= 7t 7+c = 7tan 7x+c⋅ ∫cos 8xsin 6x⋅dx= 7tan 7x+c⋅ Solve any question of Integrals with:- Patterns of problems > Was this answer helpful? 0 0 Similar questions ∫cosx(1+cosx)sinx dx=f(x)+c⇒f(x)= Medium View solution > Nettet23. mai 2024 · LHS=sin^6x+cos^6x =(sin^2x)^3+(cos^2x)^3 Using formula a^3+b^3=(a+b)^3-3ab(a+b) =(sin^2x+cos^2x)^3+3sin^2xcos^2x(sin^2x+cos^2x) =1^3+3sin^2xcos^2x*1 =1-3sin^2xcos^2x=RHS Trigonometry Science
NettetFirst note that sin6(x) + cos6(x) is a sum of two cubes (sin2(x))3 + (cos2(x))3 so it can be factored using a3 + b3 = (a + b)(a2 − ab + b2) and the Pythagorean identity to get 6 x) + cos6(x) = (sin2(x))3 + (cos2(x))3 = (sin2(x) + cos2(x))(sin4(x) − sin2(x)cos2(x) + cos4(x)) = sin4(x) − sin2(x)cos2(x) + cos4(x). Therefore,
NettetProve that sin 6x+cos 6x=1−3sin 2xcos 2x Hard Solution Verified by Toppr Consider the left hand side LHS = sin 6x+cos 6x = (sin 2x) 3+(cos 2x) 3 = (sin 2x+cos 2x) 3−3sin 2xcos 2x(sin 2x+cos 2x) = (1) 3−3sin 2xcos 2x×1 = 1−3sin 2xcos 2x =RHS Hence proved. Was this answer helpful? 0 0 Similar questions cos 22x−cos 26x=sin4xsin8x … registering iowa medicaid providerNettetThe U.S. Senate has 100 members. After a certain election, there were 18 more Democrats than Republicans, with no other parties represented. How many members … registering iot devices into dynamicsNettet2. jan. 2024 · Since the integral ranges from x = 0 to x = π, you have values of x in quadrants I and II. But in quadrant II, i.e. for x ∈ ( π 2, π), the values of cos ( x) are negative. In other words, for x ∈ ( π 2, π), we have cos ( x) = − 1 − t 2 ≠ 1 − t 2, invalidating your proposed substitution. Share Cite Follow answered Jan 2, 2024 at 21:21 zipirovich probst garden city