1. Compute 2x ecos cos sin 2.x dx e. Compute CT/4 sin* 2x cos 2x dx

Q: 3. S sin®2y · /cos2y d|

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Q: 3. Solve: cos xdx + ( 1+ | sin ædy = 0 %3D

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Q: S .2 4. sin x cosx dx

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Q: ст/2 56. " sin(2 mt/T – a) dt -

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Q: 5. S 11 sin ( 4x – 6y) cos ( 6x – 4y) dy - -

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Q: 3. у %3D In(5 — е**) 4. x = et (cos 2t + 2 sin 2t)

A: Now the first given function is ln5-e4x The second given function is x=etcos2t+2sin2t To find the…

Q: The approximation of I = = co cos (x²+5) dx using simple Simpson's rule is:

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Q: The integral / 2/ (*) cos(2*) dx takes the form k cosu du when u=2* and f (x) =x – 5. what is the…

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Q: S 5 sin(3x)e2cos(3r)dx

A: Solve integral by substitution method:-

Q: 2y у'+ — х?у? sin x %3D

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Q: Evaluate the following integral. 3 4 4 cos '3x xp. sin 3x 1 18 4 4 cos '3x dx = V1- sin 3x 18 (Type…

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Q: 5. Solve the following IVP: sin(t) y = -4 cos*(t), y(;) = cos(t) y' + 0.

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Q: e n'rdr 17 cos'x Sin 2. 4

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Q: Compute and * for the equation sin(x)e +2z²y = cos(z). 4xz COS ду

A: sinxe4xz + 2z2y = cosz∂z∂x = ?∂z∂y = ?   While calculating ∂z∂x  variable y will be considered as…

Q: 15. For fe' cos x dx, what is du? cos x e' dx - sin x dx G(e* cos x + e" sin x) + C

A: The given question can be solved as shown in step2.

Q: L[(sin2t+cos2t)² + et sinh(10t)]

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Q: Q2- Show that, 2 sin x |Jo(x cos 0) cos 0 d0

A: We have to prove sinxx=∫0π2J0(xcosθ)cosθdθ                     (1) Since we know…

Q: 2) If the secant method is used to solve the equation xe" cos(2x) = 0 in [0.7,0.8] then the vahue of…

A: The given problem is to find the approximate root x3 for the given equation using Secant method…

Q: Q1: Use the Improved Euler method to find y' = cos2x + sin3x ,0<x < 1,y(0) = 1, with h = 0.25. %3D…

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Q: 4 Using n = 8 approximate the value of cos (1+ ya) dx using Using Trapezoid rule, and Display the…

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Q: Find the Solution of differenthial Cos X

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Q: Compute S cos (x) sin²(x) dx

A: To solve the given integration problem ∫cos3(x).sin2(x) dx

Q: b) Is S cos4xv2 – sin4x dx = V(2 – sin4x)³ + C true or false? show me %3D

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Q: () = 2 + Solve the initial value problem : f'(x) = cosx + sec?x, %3D

A: As you post multiple questions we only solve first question only.

Q: The integral So (cos(3x))² sin(3x)dx under the change of variable u = cos3x becomes - ov 29 9/29 u?…

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Q: de 6) S 7) S(1 – sin? 3t) cos 3t dt 8) ſ tan (5x) sec?(5x) dx cos? e

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Q: The integral / 2' )cos(2*) dx takes the form k cosu du when u=2* and f (x) =x- 5. What is the value…

A: we use the formula ddx(ax)=ax lna

Q: x = Cịcos (At) + C2 sin (At)

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Q: dy 2y =x'cos 3x dx

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Q: Find Lt+1)²-e' sin² t + cos(t+1 ++ π)}.

A: Given :

Q: 5. Evaluate 1. S, sin(x2) dx

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Q: Consider the DE, *x' + x -(sinl+)(cos(4) 1 et what method (s) can be used to solue thio DE ? Find…

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Q: 4 sin' 2x cos 4x dx 4.

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Q: Solve the differenttal equat on 1y +dy_64=12x-50 Sin 2x dx? doe

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Q: 1 Evaluate Le' cos 2t +÷sinh 2t

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Q: 3. S sin? (2t) cos³ (2t) dt 4. ſ tan³(3x)dx

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Q: у" + 4y' + Зу %3D sin (e*)

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Q: 13 20 Find for Two only of the following: dx 1- y=xtanhx 2- y =cosh 3x- sinh²3x

A: We have to find derivative

Q: S 4 cos (x) sin (x) dx = %3D

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Q: 4. S(1 – cos 4u) du 5. S sin x cos x dx

A: 4) Given:        ∫1-cos4udu Apply linearity:    =∫1 du-∫cos4u du     =u-∫cos4u du

Q: 8. (sin°(5y)dy

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Q: The approximation of the integral (x-2) cos(x+ 1) dx using two points Gaussian quadrature formula…

A: To approximate the below integral using two points Gaussian quadrature formula. ∫−11x−22cosx+1dx

Q: 9. Evaluate ſ,(cos x – 1) dx. Tt/2 - π/2

A: topic - definite integral

Q: 2 Recall thal cos-x= arccos x, Find /2 X Sin (cos x) dx. C.

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Q: 6. In right ALMN with right angle at N, sin L 4x-0.22 and the cos M 2x+ 0.28. Determine and state…

A: Given; sin=4x-0.22 and the cosM =2x+0.28.

Q: Q: Evaluate : fe 1- cos 2ax dx. 1+ sin 2 ax 2ax

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Q: 5. Evaluate SO-2²πT ydx = 3π₁ given x = 0 - sine, y = 1- cos -0

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Q: b) Is S cos4xV2 – sin4x dx (2 – sin4x)³ + C true or false? show me

A: this is false because

Q: From the definition (Equation 4.46), construct »¡ (x) and n2(x). 1 d je(x) = (-x)' 1 d ; ne(x) = -…

A: The definition is nl(x)=-(-x)l1xddxlcos xx.   Substitute  n=1 in nl(x)=-(-x)l1xddxlcos xx.…