answer a problem

∫tan⁡x3 dx \int \sqrt[3]{\tan{x}} \, dx

#Calculus #MathProblem #Math

Note by Monojit Kamilya
7 years, 10 months ago

No vote yet
3 votes

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list
  • bulleted
  • list

1. numbered
2. list
  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai−1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 ∑i=13 \sum_{i=1}^3
\sin \theta sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Make the substitution tan⁡x=t3/2\tan x = t^{3/2}

I believe we then need to just find out ∫dt1+t3\int \frac{dt}{1+t^3}

which should be doable by partial fraction methods.

Peiyush Jain - 7 years, 10 months ago

Log in to reply

That's right. Note that 1+t3=(1+t)(1−t+t2) 1 + t^3 = (1+t)(1-t+t^2)

Also,

13(2−tt2−t+1+1t+1)=1t3+1 \frac{1}{3} \left(\frac{2-t}{t^2-t+1}+\frac{1}{t+1}\right)=\frac{1}{t^3+1}

The second part of the integral can now be solved by a simple substitution.

For the first part, use

2−tt2−t+1=−2t−42(t2−t+1)=−12(2t−1t2−t+1−3t2−t+1) \frac{2-t}{t^2-t+1}=-\frac{2 t-4}{2 \left(t^2-t+1\right)}=-\frac{1}{2} \left(\frac{2 t-1}{t^2-t+1}-\frac{3}{t^2-t+1}\right)

The first integral with 2t−1t2−t+1 \frac{2 t-1}{t^2-t+1} can be solved by a simple substitution, u=1−t+t2 u = 1 - t + t^2

For the second one, note that t2−t+1=a+(t−b)2 t^2-t+1=a+(t-b)^2 with a=34 a = \frac{3}{4} and b=12 b = \frac{1}{2}

So,

1t2−t+1=1(t−12)2+34=431(43(t−12))2+1 \frac{1}{t^2-t+1}=\frac{1}{\left(t-\frac{1}{2}\right)^2+\frac{3}{4}}=\frac{4}{3} \frac{1}{\left(\sqrt{\frac{4}{3}} \left(t-\frac{1}{2}\right)\right)^2+1}

Now make one final substitution here to arrive at the well-known ∫1z2+1 dz \int \frac{1}{z^2+1} \, dz . Your final solution looks like this:

14(−23tan⁡−1(2tan⁡(x)3+3)−23tan⁡−1(3−2tan⁡(x)3)−2log⁡(tan⁡23(x)+1)+log⁡(tan⁡23(x)+3tan⁡(x)3+1)+log⁡(tan⁡23(x)−3tan⁡(x)3+1)) \frac{1}{4} \left(-2 \sqrt{3} \tan ^{-1}\left(2 \sqrt[3]{\tan (x)}+\sqrt{3}\right)-2 \sqrt{3} \tan ^{-1}\left(\sqrt{3}-2 \sqrt[3]{\tan (x)}\right)-2 \log \left(\tan ^{\frac{2}{3}}(x)+1\right)+\log \left(\tan ^{\frac{2}{3}}(x)+\sqrt{3} \sqrt[3]{\tan (x)}+1\right)+\log \left(\tan ^{\frac{2}{3}}(x)-\sqrt{3} \sqrt[3]{\tan (x)}+1\right)\right)

Phew.

post not guaranteed to be error free

Ivan Stošić - 7 years, 10 months ago
×

Problem Loading...

Note Loading...

Set Loading...