How do you integrate int cos2x by integration by parts method?

To integrate ∫ cos(2x) dx using the integration by parts method, we start by recalling the formula for integration by parts:

∫ u dv = uv - ∫ v du

For our integration, we can choose:

• u = cos(2x) (which we will differentiate),
• dv = dx (which we will integrate).

Now we compute du and v:

• du = -2sin(2x) dx (using the chain rule),
• v = x (the integral of dv).

Substituting these into the integration by parts formula, we have:

∫ cos(2x) dx = x * cos(2x) - ∫ x * (-2sin(2x)) dx

This simplifies to:

∫ cos(2x) dx = x * cos(2x) + 2 ∫ x * sin(2x) dx

Next, we will need to apply integration by parts again to the term ∫ x * sin(2x) dx. For this, we can set:

• u = x (which we differentiate),
• dv = sin(2x) dx (which we integrate).

Now calculate du and v:

• du = dx,
• v = -0.5cos(2x) (the integral of sin(2x)).

Plugging these into our formula gives us:

∫ x * sin(2x) dx = -0.5x * cos(2x) - ∫ -0.5cos(2x) dx

This further simplifies to:

∫ x * sin(2x) dx = -0.5x * cos(2x) + 0.25sin(2x)

Now substituting this back into our earlier equation, we get:

∫ cos(2x) dx = x * cos(2x) + 2 * (-0.5x * cos(2x) + 0.25sin(2x))

Simplifying this expression, we find:

∫ cos(2x) dx = x * cos(2x) - x * cos(2x) + 0.5sin(2x) + C

Where C is the constant of integration. Thus, the final result is:

∫ cos(2x) dx = 0.5sin(2x) + C