Question: 2(b).Use tabular method to integrate Se-3x sin(9x3 + 5x2 + 3x) dx.

Q: 3.3 Inx + tanxe d 4-4x -sinte dx

A:

Q: sin 6x cos 3x dx

A:

Q: 1 -dx = /8-(2x+5)² o įsin- (2*) 2x+5 + c V8 2x+5 sin-1 + c 8, tan- ( V8 2x+5 + c o ţtan" (2*5 ) + c…

A:

Q: cos-1 3x dx 3 – 1 – 9x2 * tan-1x dx 1+ x2 4 - n/2 5 –

A:

Q: 1 -sin(2x)% = (D²-1 -1 sin(2x) 125 O ibs

A:

Q: 2cos^2(?) sin3? dx

A:

Q: 3.§ sin* 2 x sin 4 xdx

A:

Q: |5sin(4x + 5) sin(2x --) dx

A: We need to evaluate the given integral. We will use the formula  Cos(A+B)-Cos(A-B)=-2 SinA SinB

Q: 1 cos(sec-x) dx 2/V3 xVx²-1

A: Let's find.

Q: 2x sin- (x2) dæ dar 0,

A:

Q: 4x sin(4x + 5) dx 3.

A: Answer is here

Q: sin 2 2xcos 2x cos 2x dx

A:

Q: 6. f cos¹ 2 xdx

A: Given,  ∫ cos7 2xdx

Q: 4 f sin(5 – 6x) sin (6x – 4)dx -

A: Here limit are not given so we integrate only without using of limit.

Q: S sin 2x sin 4x dx 7.

A:

Q: i) Find sin 6x+ cos 4x dx. 3 ii) Evaluate || et -dx . Hence, solve 2e* dx . (* +2)" +2

A:

Q: csch (2 sin 2x)dx 1. V1-4x

A:

Q: 3- Sin4x dx S 3-Sin4x dx ノo

A:

Q: इसे १५ Sin 2X *3 tant 2X Lind ds_2

A: Given that the function is   y = 2sin2 2x × 3tan-1 2x To find the dydx

Q: 3 (4+sin-15x)° dx (8) V1-25x2

A:

Q: sin 3x dx 3-10cos3x 10.

A:

Q: sin(2x) cos(2x) dx 1+ cos(4x)

A: The complex form of integral can be converted to simple integral using integration by substitution…

Q: S sin3xsin 2x dx 3.

A:

Q: π/2 sin3 (4 ) αθ - π/2

A: To evaluate the following integrals

Q: 5sin(4x + )sin(2x-)dx

A: The given integral is, The integral cannot be solved directly and we would have to use some…

Q: 2x+5 dx = AV7- 6x - x2 + B sin - (**)- 1. Vア-6xーx2 +C

A: ∫2x+57-6x-x2

Q: csch (2 sin2x)dx 1. 1– 4x²

A:

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: sin(-4x) cos 3x dx 1. 3

A:

Q: S sin-1 3x dx

A:

Q: 비신부 4 x² sin(2x) dx

A: Given : ∫827dxx-x3

Q: 一2 sin2x dx

A:

Q: 2x sin-1(x²)dx =

A: This question is taken from the indefinite integral in which we have to choose the correct option…

Q: Evaluate the integral.

A: Given integral is ∫sin2xdx1+cos2x. Consider a=1+cos2x. Then da=-2cosxsinxdxdx=da-sin2x

Q: 1. Ssin (2x-1)cos (2x-1)dx

A: 1) Evaluate: ∫sin42x-1cos32x-1dx

Q: | sin(4x + 3)cos(3x – 5)dx -

A: We have to solve integral

Q: ( csch (2 sin 2x) dx V1-4x

A: We have to find indefinite integral

Q: S sin³ (2x)cos² (2x)dx

A:

Q: Evaluate the intergrals. ∫ ex sin2x dx   show the complete solution. step by step

A: There are exponential function and trigonometric functions in the product in the given integrand.…

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

A:

Q: cos(2x) dx 2 (sin(2x)+5)*

A:

Q: Jam sin(10x) dx

A:

Q: 6x 4. se eo* sine³x dx ‚3x

A:

Q: 5. / x² sin 3x dx

A: We have to find the indefinite integral

Q: 2cosx dx 2 1+(sinx)²

A: Please refer the attached image for complete solution.

Q: —2х Se-2x sin 3x dx

A:

Q: sin(4 + 5x) cos (5x – 4)dx

A:

Q: 4. c) Integrate x² sin(4x)dx.

A:

Q: SVx² + 10x + 25 dx. (cos4*x – sin4*x)dx dx

A: