A few integration by parts problems

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The discussion focuses on challenges faced while solving integration by parts problems, specifically three integrals involving arcsin, logarithmic, and arctan functions. The user attempted various substitutions but encountered difficulties leading to more complex integrals. Suggestions from other participants emphasized the importance of using u-substitution before applying integration by parts to simplify the problems. They advised revisiting the initial choices for u and dv to avoid getting stuck. Overall, the conversation highlights common pitfalls in integration techniques and encourages a more strategic approach to problem-solving.
kudoushinichi88
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Homework Statement



Hello. I am doing some problems on integration by parts and got stuck on the following problems. Any help would be appreciated.

i. \int \arcsin x dx

ii. \int_{0}^{1} x \ln (9+x^2) dx

iii. \int x^2 \arctan x\, dx

Homework Equations



u\,du=uv-v\,du

The Attempt at a Solution



i. I tried u= arcsin x and dv=dx, but ended up with an integral which looks worse than the original question.

ii. I used u= \ln (9+x^2)\,dx but ended up with

\int_{0}^{1} \frac{x^3}{9+x^2}\,dx
which I am stuck with.

iii. I tried u=arctan x and dv= x^2 dx but ended up with a similar looking integral in question ii, which is \int x \ln(1+x^2)\,dx
 
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You are getting stuck at a point where a simple substitution would finish it. Look at ii). Try the u-substitution u=9+x^2.
 
#1: you gave up too soon. Show us what you got.

#2: don't integrate by parts yet. Perform a u-substitution first.

#3: your choice for u and dv are fine, but I don't get that integral that's similar to #2. Again, show us what you got.
 
For 1) look at the integral:
<br /> \int 1\cdot\sin^{-1}xdx<br />
Use u=arcsin x and dv=1
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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