How do you find the integral of # (sin 2x + 1)^2#?

Answer 1

#int (sin2x+1)^2dx = 3/2 x -cos 2x -1/8sin 4x +C#

Expand the power of the binomial:

#int (sin2x+1)^2dx = int (sin^2 2x + 2 sin 2x +1)dx#

Use the linearity of the integral:

#int (sin2x+1)^2dx = int sin^2 2x dx + 2 int sin 2x dx +int dx#

Solve the integrals separately:

#int dx = x+C_1#
#2 int sin 2x dx = int sin 2x d(2x) = -cos 2x+C_2#
#int sin^2 2x dx = int (1-cos 4x)/2dx = 1/2x -1/8 sin(4x) +C_3#

And collecting the terms:

#int (sin2x+1)^2dx = 3/2 x -cos 2x -1/8sin 4x +C#

and applying trigonometric formulas to reduce to functions of the same angle:

#int (sin2x+1)^2dx = 3/2 x -cos 2x -1/4sin 2x cos 2x+C#
#int (sin2x+1)^2dx = 3/2 x -cos 2x (1/4sin 2x +1)+C#
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Answer 2

To find the integral of ((\sin(2x) + 1)^2), you can use the substitution method. Let (u = \sin(2x) + 1). Then, differentiate (u) with respect to (x) to find (du). After that, rewrite the integral in terms of (u) and (du). Finally, integrate with respect to (u) and then substitute back (u = \sin(2x) + 1) to obtain the final result.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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