# 10 Integral Problems

I wrote some problems inspired by the MIT integration bee. Enjoy ^-^

1. $$\displaystyle\int{\dfrac{dx}{x + \ln{(x^x)}}}$$

2. $\displaystyle\int{2^{x} \cdot 2^{2^{x}} \cdot 2^{2^{2^{x}}}dx}$

3. $\displaystyle\int{\sqrt{1+\sqrt{1+x}} dx }$

4. $\displaystyle\int{ \dfrac{\sin{(8x)}}{\sin{(x)}}dx }$

5. $\displaystyle\int{ \dfrac{\sqrt{1-x}}{x}dx }$

6. $\displaystyle\int{ \cos^{-1}{(x)}^2dx}$

7. $\displaystyle\int_{\frac{1}{2}}^{2} { \dfrac{dx}{x^4(1+x^3)} }$

8. $\displaystyle\int_{0}^{\pi} { \lfloor \sin{(x)} - \cos{(x)} \rfloor dx }$

9. $\displaystyle\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} { \dfrac{1-\sqrt[3]{\sin{(x)}}}{1-\sin{(x)}^2} dx}$

10. $\displaystyle\int_{0}^{1} { x^5\ln{(x)}^{10} dx}$

Note by Daniel Hinds
1 year, 6 months ago

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3.) $\int_{ }^{ }\sqrt{1+\sqrt{1+x}}dx$

$=2\int_{ }^{ }\left(t-1\right)\sqrt{t}dt\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(Took\ \sqrt{1+x}=t\right)$

$=2\left(\frac{2t^{\frac{5}{2}}}{5}-\frac{2t^{\frac{3}{2}}}{3}\right)+C$

$=\frac{4\left(1+x\right)^{\frac{5}{4}}}{5}-\frac{4\left(1+x\right)^{\frac{3}{4}}}{3}+C$

- 1 year, 6 months ago

1.) $\int_{ }^{ }\frac{1}{x\left(1+\ln x\right)}dx$

$=\ln\left(\left|1+\ln x\right|\right)+C\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(Took\ 1+\ln x\ =\ t\ \right)$

- 1 year, 6 months ago

2.) $\ln^3\left(2\right)\int_{ }^{ }\frac{\left(2^x\cdot2^{2^x}\cdot2^{2^{2^x}}\right)}{\ln^3\left(2\right)}dx$

$=\frac{\left(2^{\left(2^{2^x+1}-1\right)}\right)}{\ln^3\left(2\right)}+C\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(Took\ 2^{2^{2^x}}=t\right)$

- 1 year, 6 months ago

I think you might have made a mistake tho the answer is $\displaystyle\frac{2^{2^{2^{x}}}}{\ln^3(2)}+C$

- 2 days, 15 hours ago

5.) $2\int_{ }^{ }\frac{\cos^2t}{\sin t}dt\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(Took\ \ x\ =\ \sin^2t\right)$

$=2\left(\left(\int_{ }^{ }\csc tdt\right)-\left(\int_{ }^{ }\sin tdt\right)\right)$

$=2\ln\left(\csc t-\cot t\right)+\cos t+C$

$=2\ln\left(1-\sqrt{1-x}\right)-\ln x+\sqrt{1-x}+C$

- 1 year, 6 months ago

Excuse me, I am a student currently studying in Grade 10 and I want to master calculus. I have joined some related courses on Brilliant and I have been practicing a lot. But I am still having some difficulties in distilling the fundamental concepts in calculus. What concepts in calculus do you think are essential and fundamental? What should I pay attention on in each concept? I can only think of "limit", "continuity", "derivatives", "differentiation & integration" and a few theorems(e.g. EVT, IVT). Any comment would be a great help for me.

- 1 year, 6 months ago

Start studying this book...

- 1 year, 6 months ago

Thank you!

- 1 year, 6 months ago

Yeah Stewart calculus is bloody amazing. I use it as my primary calculus book and that's how I know calc.

- 1 year, 4 months ago

For problem 7, make the substitution $u = 1/x^3,$ $-du/3 = dx/x^4,$ to get \begin{aligned} \int_{8}^{1/8} \frac{-du/3}{1+1/u} = \frac13 \int_{1/8}^8 \frac{u \, du}{u+1} &= \frac13\left( 8-\frac18 \right) - \frac13 \int_{1/8}^8 \frac{du}{u+1} \\ &= \frac{21}8 - \frac13 (\ln(9) - \ln(9/8)) \\ &= \frac{21}8 - \frac13 \ln(8) \\ &= \frac{21}8 - \ln(2). \end{aligned}

- 1 year, 6 months ago

For problem 8, this should just be $\frac{\pi}4(-1) + \frac{\pi}4(0) + \frac{\pi}2(1) = \frac{\pi}4.$

- 1 year, 6 months ago

Solution to problem 4: WKT sin(2kx)/sin(x) = 2[cos(x) + cos(3x) +...+ cos((2k-1)x)]

Plugging k=4 gives us the integrand.

Using the R.H.S expression, the integral is 2[ sin(x) + ((sin(3x))/3) + ((sin(5x))/5) + ((sin(7x))/7) ]1😁

Really sorry for the format.

- 1 year, 6 months ago

We can use

$\frac{\sin\left(2kx\right)}{\sin x}=2\sum_{n=1}^k\cos\left(\left(2n-1\right)x\right)$

And thus, plugging $k=4$ gives us the integral to be equal to

$2\left(\sin x+\frac{\sin\left(3x\right)}{3}+\frac{\sin\left(5x\right)}{5}+\frac{\sin\left(7x\right)}{7}\right)+C$

@tc adityaa Here, I have written it in $\LaTeX$

- 1 year, 6 months ago

Problem $10$. $\int_0^1 x^5\ln x^{10}dx=10[0-\int_0^1 \dfrac{1}{x}\times \dfrac{x^6}{6}dx]=-\dfrac{5}{18}$

- 11 months, 4 weeks ago

Problem $6$. The function $\cos^{-1}(x) ^2$ doesn't have an antiderivative. To integrate the function $(\cos^{-1}x)^2$, substitute $\cos^{-1}x$ by $t$ and integrate by parts twice to obtain the value of the integral as $x(\cos^{-1}x)^2-2\sqrt {1-x^2}-2x+C$.

- 11 months, 4 weeks ago