Science:Math Exam Resources/Courses/MATH101/April 2011/Question 04 (b)

MATH101 April 2011

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Question 04 (b)
Evaluate the integral $\int x^{3}{\sqrt {1+x^{2}}}dx$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
The substitution $u=x^{2}$ almost works...

Hint 2
For an alternative solution, use a trigonometric substitution.

Hint 3
As another alternative solution, rewrite the integrand as ${\frac {x^{2}}{2}}\cdot 2x(1+x^{2})^{\frac {1}{2}}$ and apply integration by parts.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. For the change of variable $u=1+x^{2}$, we have $\displaystyle du=2x\,dx.$ So $x^{3}{\sqrt {1+x^{2}}}dx=x^{2}{\sqrt {1+x^{2}}}\,x\,dx=x^{2}{\sqrt {u}}\,{\frac {du}{2}}$ To write the remaining $x^{2}$ in terms of $u$, we note that since $u=1+x^{2}$ we have $x^{2}=u-1$. Thus ${\begin{aligned}\int x^{3}{\sqrt {1+x^{2}}}dx&=\int (u-1){\sqrt {u}}\,{\frac {du}{2}}\\&=\int (u-1)u^{1/2}\,{\frac {du}{2}}\\&={\frac {1}{2}}\int (u^{3/2}-u^{1/2})du\\&={\frac {1}{2}}\left({\frac {2}{5}}u^{5/2}-{\frac {2}{3}}u^{3/2}\right)+C\\&={\frac {1}{2}}\left({\frac {2}{5}}(1+x^{2})^{5/2}-{\frac {2}{3}}(1+x^{2})^{3/2}\right)+C\\&={\frac {1}{5}}(1+x^{2})^{5/2}-{\frac {1}{3}}(1+x^{2})^{3/2}+C.\end{aligned}}$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. This integral could be done with a trig substitution as well. We notice inside the square root that we have $1+x^{2}$. This motivates a substitution $\displaystyle {}x=\tan(\theta )$ so that we can make use of the identity $\displaystyle {}1+\tan ^{2}\theta =\sec ^{2}\theta .$ With this substitution we also have $\displaystyle {}{\textrm {d}}x=\sec ^{2}\theta {\textrm {d}}\theta .$ Putting this into our integral ${\begin{aligned}\int {}x^{3}{\sqrt {1+x^{2}}}{\textrm {d}}x&=\int {}\tan ^{3}\theta {\sqrt {\sec ^{2}\theta }}\sec ^{2}\theta {\textrm {d}}\theta \\&=\int \tan ^{3}\theta \sec ^{3}\theta {\textrm {d}}\theta \end{aligned}}$ where we have used the identity from above. This trig integral is of the form $\int \tan ^{m}\theta \sec ^{n}\theta {\textrm {d}}\theta$ with m and n odd. We want to write this in a form that we can easily integrate. If we let $\displaystyle {}u=\sec \theta$ then we have hope because we notice the derivative $\displaystyle {}\sec \theta \tan \theta$ is also there. Making this substitution with $\displaystyle {}{\textrm {d}}u=\sec \theta \tan \theta {\textrm {d}}\theta$ we get ${\begin{aligned}\int {}\tan ^{3}\theta \sec ^{3}\theta {\textrm {d}}\theta &=\int {}\tan ^{2}\theta (\tan \theta \sec \theta )\sec ^{2}\theta {\textrm {d}}\theta \\&=\int (\sec ^{2}\theta -1)\sec ^{2}\theta (\tan \theta \sec \theta ){\textrm {d}}\theta \\&=\int (u^{2}-1)u^{2}{\textrm {d}}u=\int {}u^{4}-u^{2}{\textrm {d}}u\\&={\frac {1}{5}}u^{5}-{\frac {1}{3}}u^{3}+C.\end{aligned}}$ We now have to substitute back, $u=\sec \theta =\sec \arctan {x}={\sqrt {x^{2}+1}}.$ Therefore we get $\int {}x^{3}{\sqrt {x^{2}+1}}{\textrm {d}}x={\frac {1}{5}}\left({\sqrt {x^{2}+1}}\right)^{5}-{\frac {1}{3}}\left({\sqrt {x^{2}+1}}\right)^{3}+C$

Solution 3

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We apply integration by parts with $u={\frac {x^{2}}{2}}$ and $dv=2x(1+x^{2})^{\frac {1}{2}}\,dx$. Then $du=x\,dx$ and $v={\frac {2}{3}}(1+x^{2})^{\frac {3}{2}}$. Thus ${\begin{aligned}\int x^{3}{\sqrt {1+x^{2}}}dx&=\int {\frac {x^{2}}{2}}\cdot 2x(1+x^{2})^{\frac {1}{2}}dx\\&={\frac {x^{2}}{2}}{\frac {2}{3}}(1+x^{2})^{\frac {3}{2}}-\int x{\frac {2}{3}}(1+x^{2})^{\frac {3}{2}}dx\\&={\frac {x^{2}}{2}}{\frac {2}{3}}(1+x^{2})^{\frac {3}{2}}-{\frac {1}{3}}{\frac {2}{5}}(1+x^{2})^{\frac {5}{2}}+C\\&={\frac {x^{2}}{3}}(1+x^{2})^{\frac {3}{2}}-{\frac {2}{15}}(1+x^{2})^{\frac {5}{2}}+C\end{aligned}}$ Note: The fastest way to confirm that this is the same answer as we get in the solutions above, check that they both simplify to ${\sqrt {1+x^{2}}}\left(-{\frac {2}{15}}+{\frac {1}{15}}x^{2}+{\frac {1}{5}}x^{4}\right)+C$.

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