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2026 MIT Integration Qualifier Tests

T1

$$\int_{-\pi}^{\pi} \sin^{2025}(x) \cos^{2026}(x) \, \mathrm{d}x$$

T2

$$\int e^{2026e^x + x} \,\mathrm{d}x$$

$$\begin{aligned} \int e^{2026e^x + x} \, \mathrm{d}x&=\int e^{2026e^x}e^x \, \mathrm{d}x\\ &=\int e^{2026e^x } \, \mathrm{d}e^x\\ &=\frac{e^{2026e^x}}{2026}+C \end{aligned}$$

T3

$$\int_{0}^{2026} \left\{ \frac{\lfloor x \rfloor}{3} \right\} \, \mathrm{d}x$$

$$\begin{aligned} \int_{0}^{3} \left\{ \frac{\lfloor x \rfloor}{3} \right\} \, \mathrm{d}x&=\int_{0}^{1} \left\{ \frac{0}{3}\right\} \, \mathrm{d}x+\int_{1}^{2} \left\{ \frac{1}{3}\right\} \, \mathrm{d}x+\int_{2}^{3} \left\{ \frac{2}{3}\right\} \, \mathrm{d}x\\ &=0+\frac{1}{3}+\frac{2}{3}\\ &=1 \end{aligned}$$ $$\begin{aligned} \int_{0}^{2026} \left\{ \frac{\lfloor x \rfloor}{3} \right\} \, \mathrm{d}x&=675\int_{0}^{3} \left\{ \frac{\lfloor x \rfloor}{3} \right\} \, \mathrm{d}x+0=675 \end{aligned}$$

T4

$$\int_{-1}^{1} \underbrace{\left| x + |x + |\cdots + |x + |x||\cdots|| \right|}_{2026\;x's} \,\mathrm{d}x$$

$$f_k(x)= \left \{ \begin{aligned} &0 & x<0\\ &2kx & x>0 \end{aligned} \right.$$ $$|x+|x+f_k(x)||=\left \{ \begin{aligned} &0 & x<0\\ &2x+f_k(x) & x>0 \end{aligned} \right. \\~\\ \Rightarrow f_{k+1}(x)=|x+|x+f_k(x)||$$ $$\begin{aligned} I&=\int_{-1}^{1}f_{1013}(x)\mathrm{d}x\\ &=\int_{0}^{1}2026x\mathrm{d}x\\ &=1013 \end{aligned}$$

T5

$$\int \frac{\mathrm{d}x}{\sqrt{x + 1} - \sqrt{x - 1}}$$

$$\begin{aligned} \int \frac{\mathrm{d}x}{\sqrt{x + 1} - \sqrt{x - 1}}&=\int \frac{\sqrt{x+1}+\sqrt{x-1}}{2}\mathrm{d}x\\ &=\int \frac{\sqrt{x+1}}{2}\mathrm{d}x+\int\frac{\sqrt{x-1}}{2}\mathrm{d}x\\ &=\frac{1}{3}\left((x+1)^{\frac{3}{2}}+(x-1)^{\frac{3}{2}}\right)+C \end{aligned}$$

T6

$$\int \sqrt{1 + \cosh(x)} \, \mathrm{d}x$$

$$\cosh(x)=\frac{e^x+e^{-x}}{2}$$ $$1+\cosh(x)=\frac{2+e^x+e^{-x}}{2}=\frac{\left(e^{\frac{x}{2}}+e^{-\frac{x}{2}}\right)^2}{2}=2\cosh^2\left(\frac{x}{2}\right)$$ $$\begin{aligned} I&=\int\sqrt2\cosh\left(\frac{x}{2}\right)\mathrm dx\\ &=2\sqrt 2 \sinh\left(\frac{x}{2}\right)+C \end{aligned}$$

T7

$$\int \frac{2^{\ln(x)}}{x^2} \, \mathrm{d}x$$

$$\begin{aligned} \int \frac{2^{\ln(x)}}{x^2} \, \mathrm{d}x&=-\int2^{\ln(x)}\left(-\frac{1}{x^2}\right)\mathrm{d}x\\ &\stackrel{u\gets\frac{1}{x}}{\Longrightarrow}-\int2^{-\ln(u)}\mathrm{d}u\\ &=-\int u^{-\ln2}\mathrm{d}u\\ &=\frac{1}{\ln2-1}u^{1-\ln2}+C\\ &=\frac{1}{\ln2-1}\left(\frac{1}{x}\right)^{1-\ln2}+C \end{aligned}$$

T8

$$\int_{0}^{1/2} \left( \sum_{n=2}^{\infty} x^n \right) \, \mathrm{d}x$$

$$\sum_{n=2}^{\infty} x^n =\frac{x^2}{1-x}=-\left(x+1+\frac{1}{x-1}\right)$$ $$\begin{aligned} I&=-\int_{0}^{\frac{1}{2}}\left(x+1+\frac{1}{x-1}\right)\mathrm{d}x\\ &=-\left(\frac{x^{2}}{2} + x + \ln|1-x|\right)\Bigg|_{0}^{\frac{1}{2}}\\ &=\ln2-\frac{5}{8} \end{aligned}$$

T9

$$\int x^2 \sin(x) \, \mathrm{d}x$$

$$\begin{aligned} I&=x^2\left(-\cos(x\right))-\int2x(-\cos(x))\mathrm{d}x\\ &=-x^2\cos(x)+2\int x\cos(x)\mathrm{d}x \end{aligned}$$ $$\begin{aligned} J&=x\sin(x)-\int\sin(x)\mathrm{d}x\\ &=x\sin(x)+\cos(x)+C \end{aligned}$$ $$I=-x^2\cos(x)+2x\sin(x)+2\cos(x)+C$$

T10

$$\int \frac{(x-1)^2}{2e^x + x^2 + 1} \, \mathrm{d}x$$

$$f'(x)=2e^x+2x$$ $$f(x)-f'(x)=x^2-2x+1=(x-1)^2$$ $$\begin{aligned} I&=\int\frac{f(x)-f'(x)}{f(x)}\mathrm{d}x\\ &=x-\int\frac{1}{f(x)}\mathrm{d}f(x)\\ &=x-\ln|2e^x+x^2+1|+C \end{aligned}$$

T11

$$\int_{-1}^{1} \max \left( 0, \sqrt{1 - x^2} - \frac{1}{2} \right) \, \mathrm{d}x$$

$$I=\frac{\pi}{2}-\frac{\sqrt 3}{4}-\frac{\pi}{6}=\frac{\pi}{3}-\frac{\sqrt 3}{4}$$

T12

$$\int_{0}^{1} \sqrt{x^2 + x + \sqrt{x^2 + x + \sqrt{\cdots}}} \, \mathrm{d}x$$

$$f^2(x)=f(x)+x^2+x$$ $$\begin{aligned} f^2(x)-x^2&=f(x)+x\\ (f(x)-x)(f(x)+x)&=f(x)+x \end{aligned}$$ $$f(x)-x=1\Rightarrow f(x)=x+1$$ $$I=\int_{0}^{1}(x+1)\mathrm{d}x=\frac{3}{2}$$

T13

$$\int (\cos^5(x) - 10 \cos^3(x) \sin^2(x) + 5 \cos(x) \sin^4(x)) \, \mathrm{d}x$$

$$\begin{aligned} I&=\Re\left[\int (\cos(x)+i\sin(x))^5\mathrm{d}x\right]\\ &=\Re\left[\int e^{5ix}\mathrm{d}x \right]\\ &=\Re\left[\frac{e^{5ix}}{5i}+C\right]\\ &=\Re\left[\frac{\cos(5x)+i\sin(5x)}{5i}+C\right]\\ &=\frac{\sin(5x)}{5}+C \end{aligned}$$

T14

$$\int \arctan(\sqrt{x}) \, \mathrm{d}x$$

$$\begin{aligned} \int \arctan(\sqrt{x}) \, \mathrm{d}x&\stackrel{u\gets\sqrt x}{\Longrightarrow} \int\arctan(u)\mathrm{d}u^2\\ &=2\int\arctan(u)u\mathrm{d}u \end{aligned}$$ $$\begin{aligned} J&=\arctan(u)\frac{u^2}{2}-\frac{1}{2}\int\frac{u^2}{1+u^2}\mathrm{d}u\\ &=\arctan(u)\frac{u^2}{2}-\frac{1}{2}\int\left(1-\frac{1}{1+u^2}\right)\mathrm{d}u\\ &=\arctan(u)\frac{u^2}{2}-\frac{u}{2}+\frac{\arctan(u)}{2} \end{aligned}$$ $$I=x\arctan(\sqrt x)-\sqrt x+\arctan(\sqrt x)$$

T15

$$\int_{0}^{1000} \left( \lfloor \lceil x \rceil \rfloor + \lceil \lfloor x \rfloor \rceil + \lfloor \{x\} \rfloor + \{ \lfloor x\rfloor \} + \lceil \{x\} \rceil + \{ \lceil x \rceil\} \right) \, \mathrm{d}x$$

$$I_1=500500,I_2=499500$$ $$I_3=I_4=0$$ $$I_5=1000,I_6=0$$ $$I=\sum_{i=1}^6I_i=1001000$$

T16

$$\int\sqrt{ \frac{\cos(x) \cot(x) \csc(x)}{\sin(x) \tan(x) \sec(x)}} \, \mathrm{d}x$$

$$I=\int \frac{\cos^2(x)}{\sin^2(x)}\mathrm{d}x$$ $$\cot'(x)=\frac{-\sin^2(x)-\cos^2(x)}{\sin^2(x)}=-\frac{1}{\sin^2(x)}$$ $$\begin{aligned} I&=\int\left(\frac{1}{\sin^2(x)}-1\right)\mathrm{d}x\\ &=-\cot(x)-x+C \end{aligned}$$

T17

$$\int_{-\infin}^{\infty} \frac{e^{-x^2}}{1 + e^{2x}} \, \mathrm{d}x$$

$$\begin{aligned} f(x)+f(-x)&=\frac{e^{-x^2}}{1+e^{2x}}+\frac{e^{2x}e^{-x^2}}{1+e^{2x}}\\ &=e^{-x^2} \end{aligned}$$ $$2I=\int_{-\infin}^{\infin}e^{-x^2}=\sqrt \pi$$ $$I=\frac{\sqrt \pi}{2}$$

T18

$$\int \left( \frac{\sin^2(x)}{x^2} - \frac{\sin(2x)}{x} \right) \, \mathrm{d}x$$

$$\left[\sin^2(x)\right]'=2\sin(x)\cos(x)=\sin(2x)$$ $$\begin{aligned} I&=\int\frac{\sin^2(x)-x\left[\sin^2(x)\right]'}{x^2}\mathrm{d}x\\ &=-\frac{\sin^2(x)}{x}+C \end{aligned}$$

T19

$$\int \frac{\ln(\ln(x)) \ln(\ln(\ln(x)))}{x \ln(x)} \, \mathrm{d}x$$

$$\begin{aligned} I&\stackrel{u\gets\ln(x)}{\Longrightarrow}\int\frac{\ln(u)\ln(\ln(u))}{u}\mathrm{d}u\\ &\stackrel{v\gets\ln(u)}{\Longrightarrow}\int v\ln(v)\mathrm{d}v\\ &=\frac{v^2}{2}\ln(v)-\int\frac{v^2}{2}\cdot\frac{1}{v}\mathrm{d}v\\ &=\frac{v^2}{2}\ln(v)-\frac{v^2}{4}+C\\ &=\frac{\ln^2(\ln(x))}{2}\ln(\ln(\ln(x)))-\frac{\ln^2(\ln(x))}{4}+C \end{aligned}$$

T20

$$\int_{0}^{\pi/2} \cos^2\left(\frac{\pi}{2} \cos^2\left(\frac{\pi}{2} \cos^2(x)\right)\right) \, \mathrm{d}x$$

$$\begin{aligned} J=\int_{0}^{\pi/2} \sin^2\left(\frac{\pi}{2} \cos^2\left(\frac{\pi}{2} \cos^2(x)\right)\right) \, \mathrm{d}x&=\int_{0}^{\pi/2} \cos^2\left(\frac{\pi}{2}-\frac{\pi}{2} \cos^2\left(\frac{\pi}{2} \cos^2(x)\right)\right) \, \mathrm{d}x \end{aligned}$$ $$\begin{aligned} f\left(\frac{\pi}{2}-x\right)&=\frac{\pi}{2}\cos^2\left(\frac{\pi}{2}\sin^2(x)\right)\\ &=\frac{\pi}{2}\cos^2\left(\frac{\pi}{2}-\frac{\pi}{2}\cos^2(x)\right)\\ &=\frac{\pi}{2}\sin^2\left(\frac{\pi}{2}\cos^2(x)\right) \end{aligned}$$ $$f(x)+f\left(\frac{\pi}{2}-x\right)=\frac{\pi}{2}$$ $$I=J$$ $$2I=I+J=\int_{0}^{\frac{\pi}{2}}1\mathrm{d}x=\frac{\pi}{2}\\~\\ \Rightarrow I=\frac{\pi}{4}$$