Integration by Parts - Calculus 2

1. Evaluate $\int x \arcsin(x) \,dx$.
   Answer: $\frac{x^2}{2}\arcsin(x) + \frac{1}{4}x\sqrt{1-x^2} - \frac{1}{4}\arcsin(x) + C$ Use integration by parts with $u = \arcsin(x)$ and $dv = x dx$. This gives $du = \frac{1}{\sqrt{1-x^2}} dx$ and $v = \frac{x^2}{2}$. The integral becomes $\frac{x^2}{2}\arcsin(x) - \int \frac{x^2}{2\sqrt{1-x^2}} dx$. For the remaining integral, use trigonometric substitution with $x = \sin(\theta)$, so $dx = \cos(\theta) d\theta$. The integral becomes $\frac{1}{2}\int \frac{\sin^2(\theta)}{\sqrt{1-\sin^2(\theta)}} \cos(\theta) d\theta = \frac{1}{2}\int \sin^2(\theta) d\theta = \frac{1}{4}\int (1 - \cos(2\theta)) d\theta = \frac{1}{4}(\theta - \frac{1}{2}\sin(2\theta)) = \frac{1}{4}(\theta - \sin(\theta)\cos(\theta))$. Since $\theta = \arcsin(x)$, $\sin(\theta)=x$, and $\cos(\theta)=\sqrt{1-x^2}$, this part is $\frac{1}{4}(\arcsin(x) - x\sqrt{1-x^2})$. Combining everything gives $\frac{x^2}{2}\arcsin(x) - \frac{1}{4}(\arcsin(x) - x\sqrt{1-x^2}) + C = \frac{x^2}{2}\arcsin(x) + \frac{x}{4}\sqrt{1-x^2} - \frac{1}{4}\arcsin(x) + C$.
2. If $f(x)$ is an integrable function such that $\int f(x) e^x \,dx = f(x)e^x - \int 3x^2 e^x \,dx$, what is $f(x)$?
   Answer: $x^3$ The formula for integration by parts is $\int u \,dv = uv - \int v \,du$. The given equation is $\int f(x) e^x \,dx = f(x)e^x - \int 3x^2 e^x \,dx$. We can match this to the IBP formula by identifying the parts. The original integral is $\int u \,dv$. Let's set $u=f(x)$ and $dv=e^x dx$. Then $v = \int e^x \,dx = e^x$. The right side of the IBP formula is $uv - \int v \,du = f(x)e^x - \int e^x f'(x) \,dx$. Comparing this to the given expression, we see that $\int e^x f'(x) \,dx = \int 3x^2 e^x \,dx$. This implies that $f'(x) = 3x^2$. Integrating $f'(x)$ to find $f(x)$ gives $f(x) = \int 3x^2 \,dx = x^3$ (we can ignore the constant of integration here).
3. Evaluate the definite integral $\int_{1}^{e} \sin(\ln x) \,dx$.
   Answer: $\frac{e}{2}(\sin(1) - \cos(1)) + \frac{1}{2}$ This is a cyclic integral. Let $I = \int_{1}^{e} \sin(\ln x) \,dx$. Use IBP with $u = \sin(\ln x)$ and $dv=dx$. Then $du = \frac{\cos(\ln x)}{x} dx$ and $v=x$. So, $I = [x\sin(\ln x)]_{1}^{e} - \int_{1}^{e} \cos(\ln x) \,dx = e\sin(1) - 0 - J$, where $J = \int_{1}^{e} \cos(\ln x) \,dx$. For $J$, use IBP with $u=\cos(\ln x)$ and $dv=dx$. Then $du = -\frac{\sin(\ln x)}{x} dx$ and $v=x$. So, $J = [x\cos(\ln x)]_{1}^{e} - \int_{1}^{e} -\sin(\ln x) \,dx = (e\cos(1) - 1\cos(0)) + I = e\cos(1) - 1 + I$. Substitute back: $I = e\sin(1) - (e\cos(1) - 1 + I)$. This gives $2I = e\sin(1) - e\cos(1) + 1$, so $I = \frac{e}{2}(\sin(1) - \cos(1)) + \frac{1}{2}$.
4. Let $I_n = \int (\ln x)^n \,dx$ for $n \ge 1$. Which of the following expresses a valid reduction formula for $I_n$?
   Answer: $I_n = x(\ln x)^n - n I_{n-1}$ To find a reduction formula for $I_n = \int (\ln x)^n \,dx$, we use integration by parts. Let $u = (\ln x)^n$ and $dv = dx$. Then $du = n(\ln x)^{n-1} \cdot \frac{1}{x} dx$ and $v = x$. Applying the integration by parts formula: $I_n = uv - \int v \,du = x(\ln x)^n - \int x \cdot n(\ln x)^{n-1} \cdot \frac{1}{x} \,dx$. Simplifying the integral gives $I_n = x(\ln x)^n - n \int (\ln x)^{n-1} \,dx$. Since $I_{n-1} = \int (\ln x)^{n-1} \,dx$, the reduction formula is $I_n = x(\ln x)^n - n I_{n-1}$.
5. Evaluate $\int \arctan(1/x) \,dx$.
   Answer: $x \arctan(1/x) + \frac{1}{2}\ln(x^2+1) + C$ Use integration by parts with $u = \arctan(1/x)$ and $dv = dx$. For the derivative of $u$, we use the chain rule: $du = \frac{1}{1+(1/x)^2} \cdot (-\frac{1}{x^2}) \,dx = \frac{1}{(x^2+1)/x^2} \cdot (-\frac{1}{x^2}) \,dx = -\frac{1}{x^2+1} \,dx$. We have $v=x$. Applying the formula: $\int \arctan(1/x) \,dx = x\arctan(1/x) - \int x(-\frac{1}{x^2+1}) \,dx = x\arctan(1/x) + \int \frac{x}{x^2+1} \,dx$. The remaining integral can be solved with a substitution $w=x^2+1$, $dw=2x dx$, which gives $\frac{1}{2}\ln(x^2+1)$. The final answer is $x \arctan(1/x) + \frac{1}{2}\ln(x^2+1) + C$.
6. Evaluate $\int \sec^3(x) \,dx$.
   Answer: $\frac{1}{2}(\sec(x)\tan(x) + \ln|\sec(x)+\tan(x)|) + C$ This is a classic integration by parts problem. Let $I = \int \sec^3(x) \,dx = \int \sec(x) \sec^2(x) \,dx$. Let $u = \sec(x)$ and $dv = \sec^2(x) dx$. Then $du = \sec(x)\tan(x) dx$ and $v = \tan(x)$. So, $I = \sec(x)\tan(x) - \int \tan(x) (\sec(x)\tan(x)) \,dx = \sec(x)\tan(x) - \int \sec(x)\tan^2(x) \,dx$. Using the identity $\tan^2(x) = \sec^2(x) - 1$, we get $I = \sec(x)\tan(x) - \int \sec(x)(\sec^2(x)-1) \,dx = \sec(x)\tan(x) - \int \sec^3(x) \,dx + \int \sec(x) \,dx$. This gives $I = \sec(x)\tan(x) - I + \ln|\sec(x)+\tan(x)|$. Solving for $I$ gives $2I = \sec(x)\tan(x) + \ln|\sec(x)+\tan(x)|$, so $I = \frac{1}{2}(\sec(x)\tan(x) + \ln|\sec(x)+\tan(x)|) + C$.
7. Evaluate the definite integral $\int_{0}^{1} x(1-x)^5 \,dx$.
   Answer: $\frac{1}{42}$ This can be solved efficiently with integration by parts. Let $u=x$ and $dv=(1-x)^5 dx$. Then $du=dx$ and $v = -\frac{(1-x)^6}{6}$. The integral is \[-\frac{x(1-x)^6}{6}\]_{0}^{1} - \int_{0}^{1} -\frac{(1-x)^6}{6} \,dx. The first term evaluates to $(-\frac{1(0)^6}{6}) - (-\frac{0(1)^6}{6}) = 0$. The integral part becomes $\frac{1}{6}\int_{0}^{1}(1-x)^6 \,dx = \frac{1}{6} [-\frac{(1-x)^7}{7}]_{0}^{1} = \frac{1}{6} ( (0) - (-\frac{(1)^7}{7}) ) = \frac{1}{6} (\frac{1}{7}) = \frac{1}{42}$. Alternatively, a substitution $w=1-x$ also works well.
8. Evaluate $\int x \sin(x) \cos(x) \,dx$.
   Answer: $-\frac{x}{4}\cos(2x) + \frac{1}{8}\sin(2x) + C$ First, simplify the integrand using the double angle identity $\sin(2x) = 2\sin(x)\cos(x)$, so $\sin(x)\cos(x) = \frac{1}{2}\sin(2x)$. The integral becomes $\frac{1}{2}\int x \sin(2x) \,dx$. Now use integration by parts with $u=x$ and $dv=\sin(2x)dx$. Then $du=dx$ and $v = -\frac{1}{2}\cos(2x)$. The integral is $\frac{1}{2} \left( x(-\frac{1}{2}\cos(2x)) - \int -\frac{1}{2}\cos(2x) \,dx \right) = \frac{1}{2} \left( -\frac{x}{2}\cos(2x) + \frac{1}{2} \int \cos(2x) \,dx \right)$. Integrating the final term gives $\frac{1}{2} \left( -\frac{x}{2}\cos(2x) + \frac{1}{2} \cdot \frac{1}{2}\sin(2x) \right) + C = -\frac{x}{4}\cos(2x) + \frac{1}{8}\sin(2x) + C$.
9. Applying integration by parts to the integral $\int \frac{\ln(1-x)}{x} \,dx$ with $u = \ln(1-x)$ and $dv = \frac{1}{x} \,dx$ results in which expression?
   Answer: $\ln(x)\ln(1-x) + \int \frac{\ln(x)}{1-x} \,dx$ Given the choice $u = \ln(1-x)$ and $dv = \frac{1}{x} \,dx$. We find the differentials and integrals: $du = \frac{-1}{1-x} \,dx$ and $v = \ln(x)$. Using the integration by parts formula $\int u \,dv = uv - \int v \,du$, we get: $\int \frac{\ln(1-x)}{x} \,dx = \ln(1-x)\ln(x) - \int \ln(x) \left( \frac{-1}{1-x} \right) \,dx$. Simplifying the expression gives $\ln(x)\ln(1-x) + \int \frac{\ln(x)}{1-x} \,dx$.
10. When using the tabular method for integration by parts on an integral $\int p(x)f(x) \,dx$, where $p(x)$ is a polynomial, the process terminates. This termination occurs because:
    Answer: The derivatives of the polynomial $p(x)$ eventually become zero. The tabular method is a streamlined version of repeated integration by parts. One column consists of the function chosen as $u$ (in this case, the polynomial $p(x)$) and its successive derivatives. The other column consists of the function chosen as $dv$ (in this case, $f(x)dx$) and its successive integrals. Since $p(x)$ is a polynomial of degree $n$, its $(n+1)$-th derivative will be zero. This causes the process to terminate, as the terms in the expansion will eventually become zero.
11. Find the indefinite integral $\int e^{2x} \sin(x) \,dx$.
    Answer: $\frac{1}{5}e^{2x}(2\sin(x) - \cos(x)) + C$ This is a cyclic integral requiring integration by parts twice. Let $I = \int e^{2x} \sin(x) \,dx$. First, let $u = \sin(x)$ and $dv = e^{2x} dx$. Then $du = \cos(x) dx$ and $v = \frac{1}{2}e^{2x}$. This gives $I = \frac{1}{2}e^{2x}\sin(x) - \frac{1}{2}\int e^{2x}\cos(x) \,dx$. For the second integral, let $u = \cos(x)$ and $dv = e^{2x} dx$. Then $du = -\sin(x) dx$ and $v = \frac{1}{2}e^{2x}$. This gives $\int e^{2x}\cos(x) \,dx = \frac{1}{2}e^{2x}\cos(x) + \frac{1}{2}\int e^{2x}\sin(x) \,dx = \frac{1}{2}e^{2x}\cos(x) + \frac{1}{2}I$. Substituting back: $I = \frac{1}{2}e^{2x}\sin(x) - \frac{1}{2}(\frac{1}{2}e^{2x}\cos(x) + \frac{1}{2}I) = \frac{1}{2}e^{2x}\sin(x) - \frac{1}{4}e^{2x}\cos(x) - \frac{1}{4}I$. Solving for $I$ gives $\frac{5}{4}I = \frac{1}{2}e^{2x}\sin(x) - \frac{1}{4}e^{2x}\cos(x)$, so $I = \frac{4}{5}(\frac{1}{2}e^{2x}\sin(x) - \frac{1}{4}e^{2x}\cos(x)) + C = \frac{1}{5}e^{2x}(2\sin(x) - \cos(x)) + C$.
12. Consider the integral $\int x^2 \sqrt{x+1} dx$. To evaluate this using integration by parts after making the substitution $u = x + 1$, which expression correctly represents the transformed integral?
    Answer: $\int (u-1)^2 u^{1/2} du$ When you encounter an integral with both polynomial and radical terms, substitution can simplify the expression before applying integration techniques. The key is carefully tracking how each part of the original integral transforms under the substitution. With the substitution $u = x + 1$, we have $x = u - 1$ and $dx = du$. Now we systematically replace each component of the original integral $\int x^2 \sqrt{x+1} dx$: - $x^2$ becomes $(u-1)^2$ - $\sqrt{x+1}$ becomes $\sqrt{u} = u^{1/2}$ - $dx$ becomes $du$ This gives us $\int (u-1)^2 u^{1/2} du$, which matches answer C. Looking at the incorrect options: Answer A has $u^2$ instead of $(u-1)^2$, which would mean you forgot to substitute for $x^2$ and incorrectly wrote $u^2$. Answer B incorrectly transforms $\sqrt{x+1}$ to $(u+1)^{1/2}$ when it should be $u^{1/2}$, and also has the wrong polynomial term. Answer D makes both errors from B: keeping $u+1$ in the polynomial and radical parts, suggesting confusion about which direction the substitution goes. Study tip: When making substitutions, write out the relationships clearly ($u = x + 1$ means $x = u - 1$) and substitute each piece methodically. Double-check by verifying that your new variable appears consistently throughout the transformed integral.
13. Evaluate the definite integral $\int_{1}^{e^2} \frac{\ln(x)}{\sqrt{x}} \,dx$.
    Answer: $4$ Use integration by parts with $u = \ln(x)$ and $dv = x^{-1/2}dx$. Then $du = \frac{1}{x}dx$ and $v = 2x^{1/2} = 2\sqrt{x}$. The integral is \[2\sqrt{x}\ln(x)\]_{1}^{e^2} - \int_{1}^{e^2} 2\sqrt{x} \cdot \frac{1}{x} \,dx. Evaluate the first term: $2\sqrt{e^2}\ln(e^2) - 2\sqrt{1}\ln(1) = 2e(2) - 2(1)(0) = 4e$. The integral term is $- \int_{1}^{e^2} 2x^{-1/2} \,dx = -[4x^{1/2}]_{1}^{e^2} = -[4\sqrt{x}]_{1}^{e^2} = -(4\sqrt{e^2} - 4\sqrt{1}) = -(4e - 4)$. Combining the parts gives $4e - (4e - 4) = 4$.
14. If $F(x) = \int_1^x t^2 \ln t dt$, what is $F'(x)$?
    Answer: $x^2 \ln x$ When you see a function defined as a definite integral with a variable upper limit, you're dealing with the Fundamental Theorem of Calculus, Part I. This theorem tells us that if $F(x) = \int_a^x f(t) dt$, then $F'(x) = f(x)$. To find $F'(x)$ for $F(x) = \int_1^x t^2 \ln t dt$, you simply substitute $x$ for $t$ in the integrand. The integrand is $t^2 \ln t$, so $F'(x) = x^2 \ln x$. Choice D is correct: $F'(x) = x^2 \ln x$. Choice A gives $x^2 \ln x - \frac{x^2}{2}$, which includes an unnecessary constant term. This suggests someone might have tried to evaluate the integral first, then differentiate—a much harder approach that's also incorrect here. Choices B and C, $\frac{x^3}{3} \ln x - \frac{x^3}{9}$ and $\frac{x^3}{3} \ln x - \frac{x^3}{9} + \frac{1}{9}$, come from actually computing the antiderivative of $t^2 \ln t$ using integration by parts, then differentiating. While these represent the antiderivative (with choice C including the constant from evaluating at the lower limit), the Fundamental Theorem gives us the derivative directly without needing to find the antiderivative first. Remember: When you see $\frac{d}{dx}\int_a^x f(t) dt$, don't integrate first—just replace $t$ with $x$ in the integrand. This saves time and avoids the complexity of integration by parts or other advanced techniques.
15. To evaluate $\int x^2 e^{-x} dx$ using integration by parts, what is the most efficient sequence of choices for $u$ and $dv$?
    Answer: First: $u = x^2, dv = e^{-x}dx$; Second: $u = x, dv = e^{-x}dx$ For polynomials multiplied by exponentials, choose the polynomial as u (to reduce its degree through differentiation) and the exponential as dv (since it's easy to integrate). First application: u = x², dv = $e^{-x}$dx gives du = 2x dx, v = -$e^{-x}$. This reduces the degree from x² to x. Second application on the remaining ∫x $e^{-x}$dx: u = x, dv = $e^{-x}$dx gives du = dx, v = -$e^{-x}$. Choice B would make integration increasingly difficult. Choice C has an incorrect dv in the first step. Choice D incorrectly treats $xe^{-x}$ as dv in the second step.
16. Evaluate the definite integral $\int_{0}^{\pi/2} x^2 \cos(x) \,dx$.
    Answer: $\frac{\pi^2}{4} - 2$ This integral requires two applications of integration by parts. First, let $u = x^2$ and $dv = \cos(x) dx$. Then $du = 2x dx$ and $v = \sin(x)$. The integral becomes \[x^2 \sin(x)\]_{0}^{\pi/2} - \int_{0}^{\pi/2} 2x \sin(x) \,dx = (\frac{\pi}{2})^2 \sin(\frac{\pi}{2}) - 0 - 2\int_{0}^{\pi/2} x \sin(x) \,dx = \frac{\pi^2}{4} - 2\int_{0}^{\pi/2} x \sin(x) \,dx. For the second integral, let $u = x$ and $dv = \sin(x) dx$. Then $du = dx$ and $v = -\cos(x)$. The integral is \[-x \cos(x)\]_{0}^{\pi/2} - \int_{0}^{\pi/2} -\cos(x) \,dx = 0 + \int_{0}^{\pi/2} \cos(x) \,dx = [\sin(x)]_{0}^{\pi/2} = 1. Substituting back gives $\frac{\pi^2}{4} - 2(1) = \frac{\pi^2}{4} - 2$.
17. Which of the following integrals would require integration by parts to be applied exactly three times to reach a final answer?
    Answer: $\int x^3 e^x dx$ For ∫x³eˣ dx, we need to reduce the polynomial degree from 3 to 0, requiring three applications: (1) u = x³, dv = eˣdx → remaining integral has x²; (2) u = x², dv = eˣdx → remaining integral has x; (3) u = x, dv = eˣdx → remaining integral is just eˣdx. Choice B (∫x²cos x dx) requires only two applications to reduce x² to constant. Choice C (∫x³ln x dx) requires only one application since we can use u = ln x, dv = x³dx directly. Choice D (∫eˣcos x dx) requires two applications, but this creates a cycle that leads back to the original integral, allowing us to solve algebraically.
18. For which of the following integrals is applying integration by parts with $u = x^2$ the most direct and effective first step?
    Answer: $\int x^2 \sin(x) \,dx$ Let's analyze the options. A) $\int x^2 \sin(x^3) \,dx$ is best solved with a direct u-substitution $u=x^3$. B) $\int x^3 e^{x^2} \,dx$ requires a substitution like $w=x^2$ before integration by parts. C) $\int \frac{x^2}{x^3+1} \,dx$ is also a direct u-substitution with $u=x^3+1$. D) $\int x^2 \sin(x) \,dx$ is a classic case for integration by parts. Choosing $u=x^2$ (the polynomial part) simplifies upon differentiation, and $dv = \sin(x)dx$ is easily integrated. This is the most direct and effective application of the technique among the choices.
19. When evaluating $\int e^x \sin x dx$ using integration by parts, after applying the technique twice, you obtain the equation $\int e^x \sin x dx = e^x \sin x - e^x \cos x - \int e^x \sin x dx$. What is the correct next step?
    Answer: Add $\int e^x \sin x dx$ to both sides and divide by 2 to solve for the integral This is the classic 'cyclic' integration by parts situation. When the same integral appears on both sides of the equation with opposite signs, we can solve algebraically. Adding ∫eˣsin x dx to both sides gives: 2∫eˣsin x dx = eˣsin x - eˣcos x, so ∫eˣsin x dx = (1/2)(eˣsin x - eˣcos x) + C. Choice B would create an infinite loop. Choice C would just restart the same process. Choice D is wrong because the equation shown is actually correct - this cyclic behavior is expected and useful for this type of integral.
20. Which of the following integrals can be evaluated using integration by parts, but would result in a more complicated integral if the wrong choice of $u$ and $dv$ is made?
    Answer: $\int x^2 e^x dx$ For ∫x²eˣ dx, the correct choice is u = x², dv = eˣ dx, which systematically reduces the polynomial degree. However, if we choose u = eˣ, dv = x² dx, we get v = x³/3 and the new integral becomes ∫(x³/3)eˣ dx, which is more complicated than the original. Choice A is best solved by substitution (u = x², du = 2x dx), not integration by parts. Choice B is also best solved by substitution (u = x², du = 2x dx). Choice D can be simplified using the power rule: ∫x³ · $x^{1/2}$ dx = ∫$x^{7/2}$ dx, which doesn't require integration by parts at all.

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