5х, 5x sin°(cos* () dx 5 4 6 6

Q: cos2x dx 1 sin 2x- 16 12

A: The objective is to find  the integration.

Q: 21 · 8* cos (8*) dx =

A: ∫21*8xcos(8x)dxSubstitute u=8x→dudx=ln(8)*8x→dx=1ln(8)*8xdu=21ln(8)∫cos(u)du

Q: sin(cos¬1/2x²) = v3x? sO:

A: Given:   SinCos-1 2x2=3x2   To find out: x=?

Q: 4. x = 3 + 4 sine, y = 3cos0 - 4

A: Given: x=3+4sinθ, y=3cosθ-4 dxdθ=ddθ3+4sinθ = 4cosθ dydθ=ddθ3cosθ-4 = -3sinθ

Q: 16. Calculate (sin 0 , 0 , cos 20) d0.

A: ∫0π< sinθ,θ,cos 2θ>dθ

Q: 11. cos4x-cos2 dx sin4x-cos2x

A:

Q: V3 sin(2r) = - Cos(2r)

A:

Q: y = 2 sin 3x 1+3 cos2x

A:

Q: 2ct-3) cos Cも-3) yct) = sin 2t cos 3t

A:

Q: 4) Sec CA-B) = cos CA+B) CoS A - Sin²B

A:

Q: symbolnc cos(6x) cos(11x) dx = cos(11) sin(6x) + C ncorrect

A:

Q: Evaluate the integral

A: Let the given integral be y - y = ∫6sin2x1+cos4xdxy = 6∫sin(2x)1+cos2x2dx…

Q: 6e(6 sin 3x) dx sec 3x

A: Inverse of sec3x is cos3x Cos3x=(1/Sec3x) Putting in the integration problem

Q: 1. [(10 cos2x - 5 csc2x +4 tan2x + 8 sin2x)dx

A: Integration properties help to find the integration of the given function. Integration property for…

Q: 12) (sin (4x - 9) dx

A: Given function : sin (4x -9) To find : Integral of the function.

Q: Evaluating √√9-x² dx using trigonometric substitution will (A) 18 18 [*/2 cos²0d0 0 **/2 B 0 Cab…

A:

Q: Cos(11°+x)=-sin(2x+6)

A: x

Q: ∫((cos3x)/(1-sinx))dx

A: We have to find the integration.

Q: sin 2x (1 - 2 cos 2x)? dx 6.

A:

Q: 1- cosx dx A -2/2cos- +C B 2/2cos- +c -2/2 sin-+C 2 D 2/2 sin- +C

A:

Q: 3nt 3nt s = sin 2 + cos 2 %3D

A: # we are entitled to solve one question at a time , please resubmit the other question if you wish…

Q: I+ Sin2X t Cos 2x COSX

A: Given,           1+sin2x+cos2xcosx

Q: I = sin³x cos³x dx

A: Given the integration : ∫sin3 xcos3 xdx

Q: 6. cos-1 Vx dx

A:

Q: 1 dx 6- cos²xsin²x

A: use trigonometric identity to prove the integral

Q: 8 csc? 4x cos(cot(4x)) dx sin² (cot(4x))

A: Here, I will be using method of substitution to solve this integral. You can assume:…

Q: sin(7x 2) dx cos2 (7x 2)

A: Let u=cos(7x+2)differentiate both sides w.r.t. x.Use chain rule for cos(7x+2)From there we get du/-7…

Q: 2 + 3 sin 2x The value of 3x dx equals 1+ cos 2x

A:

Q: Cos^3(3x) dx

A:

Q: 345 cos dx 1-cos24x+sin? 4x

A:

Q: | sin(4x + 3)cos(3x – 5)dx -

A: We have to solve integral

Q: Find the general solution to the equation/3 cos(2x) + 3 sin(2x) = 0.

A:

Q: sin3xdx Cos4x

A:

Q: cOs Xx dx 4 + sin² x 9.

A: from the given data

Q: cos° (3x) sin (3x) dr

A: Given data:     The given integral expression is: .   Assume cos(3x)=t, and differentiate it both…

Q: Solve 1 – sin (A)+1= 2 cos(A) over 0, 27) -

A:

Q: у%3D sin(4x — 2) cos (4x + 2) || -

A: topic - derivative

Q: 3. sin(2x - 6) = cos(4x + 18) %3D

A:

Q: 4). | (5 sin 2x) dx = 2

A:

Q: 2. cos?(2x) – sin²(2x) = = Cos 4x

A: We have to proof

Q: CoS 2n du.

A:

Q: sin 0 cos 0 d0. Jo 0,

A: Here we to solve definite integral.

Q: 10. 4 COS dx 2

A: the given integral is: ∫cos4x2dx we have to evaluate the given integral.

Q: 24) y = Sin' 2x.Cos2x (ans

A:

Q: cos8x -dx sin4x

A: We use some trignometric identites.  Sin2 x + cos2 x =1 Sin(2x) = 2cos(x)sin(x)

Q: -2% sin 4x dx

A:

Q: 5TT i sin' cos*dx 4x. 4x, 8. 5

A:

Q: 2. 6 cos 2x dx

A: To evaluate the given integral.

Q: 5. sin'(21) cos°(21) dt

A: