Q1-6 * 2 points 1 -sin(2x): - sin(2x) -125 (D²-1)'

Q: 7x cos х -7x - lim [sin(-2x)sin(5x)sin(7x)] 2x3cos(x) lim 5x2 х>0

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Q: -х- 2, х<-1 -1, f(x) =. -1sx< 0 at x = -1, 1, 0 sin(2tX), 0<x<1 X, 1s x

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Q: A) By using simple Iterative method, find one root of the equation sin x = Use 3D. %3D (x2- In x)3

A: We have sinx=1x2-lnx3 Now, if we need to find a solution of fx=0 then we write fx=0 in the form x=gx…

Q: 5. tan (x)dx 1 -dx cos² (x) sin () 6.

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Q: s sin² (3x) cos³ (3x) dx Hint: cos²x = 1 - sin² x 07 sin²x = 1 - cos² x

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Q: solve using reduction of order (D^2+1)y=tanx

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Q: Solve 4 sin 2x S1₁= 8 4 -1 -¹x³ + x²-x² +2 dx

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Q: Suppose that 3x² - y² = 1² and y=tsin 0. Find (3). = sin 0 y (1) ¹ (2) and (2) 0 X X

A: Let's find partial derivative of given two equation.

Q: Solve for 0 <x< 2n. 4sin'x - sinx = 0 %3D

A: Solution of trigonometric equation

Q: Verify the identity.

A: Given: arctanx=12arctan2x1-x2. The interval given is -1&lt;x&lt;1. From the given interval, let…

Q: 3 cos(x) sin(4x) 3. lim x→0 2x

A: Solution:

Q: 4. sin2xcosx + sin x 1 xE[-27, Zn]

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Q: FIN 2 dx

A: Note:- In given problem, according to written way sin2x is multiplied by 3x+1. If you want solution…

Q: Q No 4) Integrate S cosec³xdx . AND also Find F sin10xdx

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Q: -1 4.. у %3D 2а3 sinh "(6ӕ — 5) 2х-3 coth (x + 2) -1 5. у - 3 6x

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Q: 3. sin(3x) = 3s in x – 4sin³ x

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Q: Find y': COS X y = 1-sin x O y' = sin² -sin x-cos² x (1-sin x)²

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Q: | sin x (3cosx)dx 2: VZ 3x dx V1- x4 3:

A: SOLUTION-

Q: 1) y = x3 + sin x² + sin-1 5x 2) y = 32x + In (sin x²) + In(sin¯' 5x)

A: some functions are given , we have to find its derivative

Q: Evaluate the integrals

A: Given integral: ∫π6π2sin2x2sinxdx

Q: 5. 4 (secx-1)(sec² x+secx+1) Sª cos² 2x-2 cos 2x+1 -dx

A: Indefinite integral is the process through which one can determine the function which on…

Q: 2. 2. COS X Sinx lim 4x-X

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Q: 2) y - • sin 0 +2 3元 2元 5元 3元 元 元2

A: The given function,       y=12.sin θ+2

Q: 2cos(2x)sin(2x)-sin(2x)=0 Interval [0,2pie)

A: Given that the trigonometric equation, Solve the equation for x,

Q: 2. Solve 3shn-sinx-2= over the interval [0,27). 2 sın'x-sinx-|:0

A: The objective is to solve the above expression.

Q: 4. sin 5. y = x ArcCC 1 + v1-x² 2+X 6. y = Arccot x + Arctan 1-2x >| ×

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Q: 2. f(-7sinx + 3cosx)dx (-7sin 3. √(3x − 2 + sin x) dx TE 4. (3 sec² x + 4 csc² x) dx

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Q: 1-sin x dx 1-cos x 2

A: Introduction :

Q: 4. y = √esinx + 2 5. y = 4x In 4x 6. xe = sin 5x - sin 3y

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Q: 3. sin 8x cos 8x dx s 2x sin 2x dx cos

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Q: 4x 3-limx-0 tan5x sin(cos y) 4-lim, y- 2

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Q: i) etan-¹x 12 1 + x² dx ii) ii) sin sin5 x dx

A: Solve the following

Q: 6 lim Sinx ex -1

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Q: Exercises to solve: 1. fx² In(x + 1) dx 2. (x2 + x – 2).cos4x. dx 3. Se-*sinxdx 4. S sinh2x.cosh5x…

A: “Since you have asked multiple questions, we will solve the first question for you. If you want any…

Q: 4 cos °x sin x dx 3x sin: 1/3

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Q: 1. xý + 3y += sin x v2

A: Given :

Q: Find fif f'(0) = 2 sin 0-sec²0

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Q: 1. f(x) = cos 3x2. y = - tan 12x3. f(x) = sec x? 4. y = sin(2x² + 5)5. y = 12 csc6. f(x) = sin(x +…

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Q: dx 2 sin x - cOS X |tan+ 2 + V5| In 5 tan; + 2– V5 |tan+2-V5 + C + 2+ V5 2. + C In Option 1 O Option…

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Q: 9. sec(1- x) tan(1 – x) dx

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Q: 3T 2cosx dx 1+(sinx)2 2 a)2 tan-'sinx|| b) 2 In sinx| c)2 In(1 + (sinx)² 2

A: We know that integration of 1/(a2+x2) is (1/a) tan-1(x/a) +c. Here' c 'is constant. Here i attached…

Q: when

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Q: Cos x dx In x (sin x) () - dæ. (In x)²

A: We have to find the value of the integral : ∫cos(x)ln(x)dx-∫sin(x)1xln(x)2dx

Q: Find a root of  x- sin(x) - (1/2)= 0 correct to four decimal places by Secant Method

A: The given function is: x-sinx-12=0 Let, fx=x-sinx-12 Here, f0=-0.5f1=-0.3415f2=0.5907

Q: 3 8. y = sec Зx) 9sec (2 – 3x)tan´(2 – 3x) B. - 9sec (2 - 3x) tan tan (2 3 С. 9 sec' (2 — 3х) tan…

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Q: 5-7 -1 Using quotient rule, show-that d [cotx] dx 2. 1+x

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Q: sec² x 12. S 1+tanx

A: here we use       ddxtanx=sec2x see below the calculation and answer s

Q: 3. SIX f- - dx x in x√√(inx)2-9

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Q: 6) csch-1x = In Vx2+1 (+ sinh-11 %3D |x|

A: 6. First  we will find relation between csch-1x and sinh-11x. From Hyperbolic trigonometric…