Let's analyze each integral for which values of \( t \) it can be evaluated using standard calculus methods (substitution, integration by parts, trigonometric identities, partial fractions): --- ### (a) \( \int \sin(x^t) dx \) - **If \( t = 1 \):** \( \sin(x) \) can be integrated directly. - **If \( t = -1 \):** \( \sin(1/x) \) doesn't have an elementary antiderivative. - **If \( t = n \) or \( t = 1/n \):** For \( t = 2 \), \( \sin(x^2) \) does not have an elementary antiderivative. Substitution doesn't help for general \( t \). - **If \( t = -n \) or \( t = -1/n \):** Similar problem. **Answer:** **None of the provided options give an integral that can be evaluated in terms of elementary functions.** **(If \( t = 1 \) were an option, it would work, but it isn't listed.)** --- ### (b) \( \int x^t \arctan(x) dx \) - Here, integration by parts is possible for any \( t \), since we can set \( u = \arctan(x) \), \( dv = x^t dx \). - The antiderivative is not always elementary, but for any \( t \), this is a standard integration by parts situation. **Answer:** **All values:** A. \( t = n \), B. \( t = 1/n \), C. \( t = -1 \), D. \( t = -n \), E. \( t = -1/n \) (You can always use integration by parts, even if the result is not always "nice".) --- ### (c) \( \int x^t e^{x} dx \) - This is a classic case for integration by parts for any \( t \). - For integer \( t \), repeated integration by parts will eventually eliminate the \( x^t \) term. - For non-integer \( t \), the result is in terms of the gamma function or an infinite series, but the method still works. **Answer:** **All values:** A. \( t = n \), B. \( t = 1/n \), C. \( t = -1 \), D. \( t = -n \), E. \( t = -1/n \) --- ### (d) \( \int \ln(x^t) dx \) - \( \ln(x^t) = t \ln(x) \). So the integral becomes \( t \int \ln(x) dx \), which is always doable for any \( t \). **Answer:** **All values:** A. \( t = n \), B. \( t = 1/n \), C. \( t = -1 \), D. \( t = -n \), E. \( t = -1/n \) --- ### (e) \( \int x^t \ln(x) dx \) - Integration by parts works for any \( t \). **Answer:** **All values:** A. \( t = n \), B. \( t = 1/n \), C. \( t = -1 \), D. \( t = -n \), E. \( t = -1/n \) --- ## **Summary Table:** | | t = n | t = 1/n | t = -1 | t = -n | t = -1/n | |-----------|-------|---------|--------|--------|----------| | (a) | | | | | | | (b) | ✓ | ✓ | ✓ | ✓ | ✓ | | (c) | ✓ | ✓ | ✓ | ✓ | ✓ | | (d) | ✓ | ✓ | ✓ | ✓ | ✓ | | (e) | ✓ | ✓ | ✓ | ✓ | ✓ | - (a): None - (b)-(e): All values Let me know if you need solutions or explanations for any specific integral!

Let's analyze each integral carefully, perform the necessary calculus steps, and determine the values of \( t \) for which the integral can be evaluated using methods such as substitution, integration by parts, trigonometric identities, or partial fractions. --- ### **(a) \( \displaystyle \int \sin(x^{t})\, dx \)** **Goal:** Find for which \( t \) this integral can be evaluated with learned methods. --- #### Step 1: Consider substitution - The integrand is \( \sin(x^{t}) \). - A natural substitution is \( u = x^{t} \). **Compute:** \[ du = t x^{t-1} dx \Rightarrow dx = \frac{du}{t x^{t-1}} \] But note that: \[ x^{t-1} = (x^{t}) \cdot x^{-1} = u \cdot x^{-1} \] Thus, \[ dx = \frac{du}{t u x^{-1}} = \frac{x}{t u} du \] But \( x = u^{1/t} \), so \[ dx = \frac{u^{1/t}}{t u} du = \frac{1}{t} u^{\frac{1}{t} - 1} du \] The integral becomes: \[ \int \sin(u) \cdot dx = \int \sin(u) \cdot \frac{1}{t} u^{\frac{1}{t} - 1} du \] This is: \[ \frac{1}{t} \int u^{\frac{1}{t} - 1} \sin(u) du \] --- #### Step 2: Can this integral be evaluated? - The integral \( \int u^{\alpha} \sin(u) du \) is **not** elementary unless \( \alpha \) is an integer **and** related to the form of derivatives of sine functions. - In general, it's a **non-elementary integral** unless \( \alpha \) is an integer, which occurs when \( \frac{1}{t} - 1 \in \mathbb{Z} \). Solve for \( t \): \[ \frac{1}{t} - 1 = m \in \mathbb{Z} \Rightarrow \frac{1}{t} = m + 1 \Rightarrow t = \frac{1}{m + 1} \] - For example, if \( m = 0 \Rightarrow t=1 \), then the integral simplifies to: \[ \frac{1}{1} \int u^{0} \sin(u) du = \int \sin(u) du = -\cos(u) + C \] which is elementary, and the original integral becomes: \[ \int \sin(x^{1}) dx = -\cos(x) + C \] - Similarly, for other integer \( m \), the integral reduces to a form involving elementary functions via repeated integration by parts or known integrals. --- ### **Final conclusion for (a):** - The integral can be evaluated explicitly when \( t = 1/(m+1) \) with \( m \in \mathbb{Z} \). - The simplest case is \( t=1 \). **Given options:** - \( t=1 \) (Option B): **Yes**, integral reduces to elementary form. --- ### **Answer for (a):** **Values of \( t \):** **\( t=1 \)** **Corresponds to:** Option **B**: \( t= \% \) is not specified numerically in options, but based on the options in the image, the correct choice is **B. \( t = \% \)** (assuming they mean \( t=1 \)). *(In the given options, B corresponds to \( t= \% \); this likely indicates \( t=1 \).)* --- ### **(b) \( \displaystyle \int x^{t} \arctan(x) dx \)** --- #### Step 1: Use **Integration by Parts**: Set: \[ u = \arctan(x) \Rightarrow du = \frac{1}{1 + x^{2}} dx \] \[ dv = x^{t} dx \Rightarrow v = \frac{x^{t+1}}{t+1} \quad \text{(for } t \neq -1) \] **Note:** For \( t = -1 \), \( v = \ln|x| \). --- #### Step 2: Write the integral: \[ \int x^{t} \arctan(x) dx = u v - \int v du \] \[ = \arctan(x) \cdot \frac{x^{t+1}}{t+1} - \int \frac{x^{t+1}}{t+1} \cdot \frac{1}{1 + x^{2}} dx \] --- #### Step 3: Simplify the remaining integral: \[ I = \frac{1}{t+1} \int \frac{x^{t+1}}{1 + x^{2}} dx \] - For the integral to be elementary, the integrand should be manageable. --- #### Step 4: Consider substitution: \[ w = x^{2} \Rightarrow dw = 2x dx \] But the numerator involves \( x^{t+1} \). Let's analyze: \[ x^{t+1} = x^{t} \cdot x = x^{t} \cdot \sqrt{w} \] - For \( t \) such that \( x^{t} \) can be expressed in terms of \( w \), this can be simplified. --- ### **Key observation:** - When \( t = -1 \), \( x^{t} = x^{-1} \), and the integrals are manageable. - For \( t = n \) (integer), the integral involves polynomial powers. - For general \( t \), the integral can be handled via substitution if \( t \) is an integer or such that the powers cancel or produce manageable forms. --- ### **Final conclusion:** - **The integral can be evaluated explicitly for:** - \( t = -1 \), because then, \[ I = \frac{1}{0} \times \text{(undefined)} \] but in the case \( t = -1 \), the integral reduces to: \[ \int x^{-1} \arctan(x) dx \] which is manageable via integration by parts, as shown. - For other \( t \), the integral is generally complicated unless \( t \) is an integer, especially \( t=n \). --- ### **Answer for (b):** **Values of \( t \):** - \( t = n \) (any integer) - \( t = -1 \) **In options:** - **Option A: \( t = n \)** - **Option C: \( t = -1 \)** --- ### **(c) \( \displaystyle \int x^{t} e^{x} dx \)** --- #### Step 1: Use **Integration by parts**: Set: \[ u = x^{t} \Rightarrow du = t x^{t-1} dx \] \[ dv = e^{x} dx \Rightarrow v= e^{x} \] Integral: \[ \int x^{t} e^{x} dx = x^{t} e^{x} - t \int x^{t-1} e^{x} dx \] This is a recursive relation decreasing the power \( t \) by 1 each time. --- #### Step 2: When can this be evaluated? - For **integer \( t \)**, repeated application of integration by parts reduces the power to zero, leading to elementary integrals involving \( e^{x} \). - For **non-integer \( t \)**, the recursion involves non-polynomial powers, but the method still applies. --- ### **Conclusion:** The method **works for all \( t \)**, whether integer or fractional, since the recurrence is always valid. --- ### **Answer for (c):** **Values of \( t \):** All \( t \) (any real number). --- ### **(d) \( \displaystyle \int \ln(x^{t}) dx \)** Note: \[ \ln(x^{t}) = t \ln x \] So, \[ \int \ln(x^{t}) dx = t \int \ln x dx \] which is a standard integral: \[ \int \ln x dx = x \ln x - x + C \] --- ### **Final conclusion:** - For any \( t \), the integral reduces to a multiple of a basic integral involving \( \ln x \), which is elementary. --- ### **Answer for (d):** **Values of \( t \):** All \( t \). --- ### **(e) \( \displaystyle \int x^{t} \ln x dx \)** --- #### Step 1: Use **Integration by parts**: Set: \[ u = \ln x \Rightarrow du = \frac{1}{x} dx \] \[ dv = x^{t} dx \Rightarrow v= \frac{x^{t+1}}{t+1} \quad (t \neq -1) \] Integral: \[ \int x^{t} \ln x dx = \frac{x^{t+1}}{t+1} \ln x - \int \frac{x^{t+1}}{t+1} \cdot \frac{1}{x} dx \] Simplify: \[ = \frac{x^{t+1}}{t+1} \ln x - \frac{1}{t+1} \int x^{t} dx \] which is elementary for all \( t \neq -1 \): \[ \int x^{t} dx = \frac{x^{t+1}}{t+1} \] --- ### **Conclusion:** - The integral is elementary for **all \( t \neq -1 \)**, and by continuity, it can be evaluated using elementary functions for **any \( t \)** (except \( t = -1 \), where \( v=\ln x \)). --- ### **Final answer:** - **All \( t \),** with \( t \neq -1 \). --- ## **Summary of Results:** | Part | Values of \( t \) where integral can be evaluated | Explanation | |---------|---------------------------------------------------------|--------------| | **(a)** | \( t=1 \) (Option B) | Substitution simplifies the integral to elementary functions when \( t=1 \). | | **(b)** | \( t=n \), \( t=-1 \) | Integration by parts works for integers and \( t=-1 \). | | **(c)** | All \( t \) | Recursive integration by parts always applicable. | | **(d)** | All \( t \) | \( \int \ln x dx \) is elementary; scaled by \( t \). | | **(e)** | All \( t \neq -1 \) | Integration by parts yields elementary functions. | --- ### **Final notes:** - The key to each part is recognizing the applicable method (substitution, parts, identities). - For some integrals, the method applies for all \( t \), while others require specific values or conditions. --- **If you'd like, I can provide step-by-step calculations for each integral in more detail.**