Question

$$ \text { use integration by parts to evaluate each integral. } $$ $$ \int(t-3) \cos (t-3) d t $$

Short answer

\((t-3) \sin(t-3) + \cos(t-3) + C\) is the solution.

Step-by-step solution

Choose Functions for Integration by Parts

Integration by parts formula is given by \[ \int u \, dv = uv - \int v \, du \]For the integral \( \int(t-3) \cos (t-3) \; dt \), let's choose \( u = (t-3) \) (so \( du = dt \)) and \( dv = \cos(t-3) \; dt \) (so \( v = \int \cos(t-3) \; dt = \sin(t-3) \)).

Apply Integration by Parts Formula

Substitute the selected \( u \), \( dv \), \( v \), and \( du \) into the integration by parts formula:\[ \int(t-3) \cos (t-3) \; dt = (t-3) \sin(t-3) - \int \sin(t-3) \; dt \]

Integrate the Remaining Integral

Calculate the integral of \( \sin(t-3) \). The indefinite integral is:\[ \int \sin(t-3) \; dt = -\cos(t-3) + C \] where \( C \) is an arbitrary constant.

Simplify and Combine Integrals

Substitute the result from Step 3 back into the expression obtained in Step 2:\[ (t-3) \sin(t-3) - ( -\cos(t-3) + C ) = (t-3) \sin(t-3) + \cos(t-3) - C \]

Finalize the Solution

Combine constants and finalize the expression. Since \( C \) is arbitrary, the integrated result of the initial integral is:\[ \boxed{(t-3) \sin(t-3) + \cos(t-3) + C} \]

Key concepts

Definite and Indefinite Integrals

Integrals are a core part of calculus, used to compute areas, volumes, central points and many other useful things. When we talk about integral calculus, we usually deal with two types of integrals: definite and indefinite.

• Definite Integrals: These are integrals with upper and lower bounds. They're represented as \( \int_a^b f(x) \, dx \) and calculate the net area under the curve \( f(x) \) from \( x = a \) to \( x = b \).
• Indefinite Integrals: Unlike definite integrals, indefinite integrals don't have bounds. They're represented simply as \( \int f(x) \, dx \), and the result includes a constant \( C \). This constant is essential in integration as it accounts for any vertical shifts of the antiderivative.

In our problem, we're dealing with an indefinite integral \( \int (t-3) \cos(t-3) \; dt \). The solution involves finding the antiderivative and includes the constant \( C \). This

• makes indefinite integrals particularly general because any number can be substituted for \( C \) to get different valid antiderivatives.

Trigonometric Integrals

Trigonometric integrals involve trigonometric functions such as sine, cosine, tangent, etc. These types of integrals often require specific techniques or identities to simplify and integrate them efficiently.

• Trigonometric identities, like \( \sin^2(x) + \cos^2(x) = 1 \), can be very helpful in simplifying trigonometric integrals.
• For the integral \( \cos(t-3) \), the operation \( \int \cos(u) \, du \) results in \( \sin(u) + C \).

In the step-by-step solution, we encounter \( \int \sin(t-3) \; dt \), which results in \( -\cos(t-3) + C \). Here,

• the integration of sine functions inherently involves a negative sign, highlighting the importance of understanding trigonometric integral properties.

These expressions emphasize understanding the integral properties of sine and cosine, which are fundamental in solving problems involving trigonometric functions.

Integration Techniques

Choosing the appropriate technique for integration can greatly simplify the process. One popular method is integration by parts, which is useful when dealing with products of functions. The formula is:

\[ \int u \, dv = uv - \int v \, du \]

• This method is often employed when one part of the product is easy to differentiate and the other is easy to integrate.
• In the exercise, \( u = (t-3) \) makes \( du = dt \), and \( dv = \cos(t-3) \, dt \), which was integrated to get \( v = \sin(t-3) \).

Integration by parts sometimes requires the process to be repeated or combined with other techniques to fully solve the integral. This technique can turn a seemingly complex integral into a more manageable form by decomposing it into several simpler integrals. After applying the integration by parts formula, simplifying and combining results are crucial to achieving the final solution, as shown in the exercise where upon integrating \( \sin(t-3) \), it yielded the final result as a combination of terms plus the constant \( C \).
This exemplifies the strength of integration by parts in tackling functions that are otherwise difficult to integrate directly.