# Failing a basic integration exercise; where did I go wrong?

(This is a basic calculus exercise gone wrong where I need some feedback to get forward.)

I've attempted to calculate an integral by first integrating it by parts and then by substituting. The result I got is not correct though. Can I get a hint about where started to make mistakes?

Here are my steps:

$$\int \sqrt{x}\cdot\sin\sqrt{x}\,dx$$

1. Applying integration by parts formula

$$\int uv' \, dx = uv - \int u'v \, dx$$

where $$u=\sqrt{x}$$, $$v'=\sin\sqrt{x}$$, $$u'=\frac{1}{2\sqrt{x}}$$ and $$v=-\cos\sqrt{x}$$, resulting in:

$$\int \sqrt{x}\cdot\sin\sqrt{x}\,dx = \sqrt{x} \cdot -\cos\sqrt{x} - \int \frac{1}{2\sqrt{x}} \cdot -\cos\sqrt{x} \, dx$$

2. Next, I tried substituting $$x$$ with $$g(t)=t^2$$ in the remaining integral, i. e. replacing each $$x$$ in the integral with $$t^2$$ and the ending $$dx$$ with $$g'(t)=2t\,dt$$. I would later bring back the $$x$$ by substituting $$t=\sqrt{x}$$ after integration. Continuing from where we left by substituting:

$$\sqrt{x} \cdot -\cos\sqrt{x} - \int \frac{1}{2\sqrt{x}} \cdot -\cos\sqrt{x} \, dx$$ $$\Longrightarrow \sqrt{x} \cdot -\cos\sqrt{x} - \int \frac{1}{2\sqrt{t^2}} \cdot -\cos\sqrt{t^2}\cdot2t \, dt$$

3. Then I pulled out the constant multipliers from the integral:

$$\sqrt{x} \cdot -\cos\sqrt{x} - \frac{1}{2} \cdot -1 \cdot 2 \int \frac{1}{\sqrt{t^2}} \cdot \cos\sqrt{t^2}\cdot t \, dt$$

which turned out to eliminate each other (resulting in just $$\cdot1$$), so we end up with:

$$\sqrt{x} \cdot -\cos\sqrt{x} \cdot \int \frac{1}{\sqrt{t^2}} \cdot \cos\sqrt{t^2}\cdot t \, dt$$

4. Reducing the integrand by reducing $$\frac{1}{\sqrt{t^2}}\Rightarrow\frac{1}{t}$$, which again is eliminated by multiplying with the integrand's $$t$$, and reducing $$\cos\sqrt{t^2}\Rightarrow\cos t$$. Therefore resulting in:

$$\sqrt{x} \cdot -\cos\sqrt{x} \cdot \int \cos t \, dt$$

where the integral can be solved as $$\int \cos t \, dt = \sin t + C$$. So now we are at:

$$\sqrt{x} \cdot -\cos\sqrt{x} \cdot \sin t + C$$

The correct answer however is $$\int \sqrt{x} \sin\sqrt{x} \, dx = 4 \sqrt{x} \sin\sqrt{x} - 2 (x - 2) \cos\sqrt{x} + C$$

So something somewhere in my process went horribly wrong. What?

• Welcome to the MSE. – Sebastiano Jan 21 at 21:53
• Upvoted because excellent use of MathJax for a first-time user, and lots of effort has gone into answering the question. Other first-time users- take note! – Adam Rubinson Jan 21 at 21:57
• @AdamRubinson Approved also by me :-)..I am an user of TeX.SE. – Sebastiano Jan 21 at 22:00

$$v'=\sin\sqrt{x}\$$ does not imply $$\ v=-\cos\sqrt{x}$$.

By the chain rule, $$\frac{d}{dx}(-\cos(x^\frac{1}{2})) = \frac12x^{-\frac12} \times \sin\left(x^\frac12\right) \neq \sin\sqrt{x}$$.

In fact, to find $$v$$, which equals $$\int \sin\sqrt{x}\ dx$$, you have to use a substitution like $$u^2 = x.$$ Then by integrating by parts, you should get that $$v = 2\sin(\sqrt{x}) - 2\sqrt{x}\cos(\sqrt{x}).$$

You can continue down this path, but it is a bit ugly, so let's see if there's a simpler overall approach. The original integral is $$\int \sqrt{x}\cdot\sin\sqrt{x}\,dx$$. It's best to just start with the substitution $$u=\sqrt{x}\quad (^*)$$.

Then the integral becomes $$2\int u^2 \sin(u) du$$, which is much nicer to work with. I think you solve this by integrating by parts twice.

$$(^*)$$ I originally used the substitution $$\ x = u^2\$$ but I think this is inaccurate because $$\sqrt{u^2} = |u|,\$$ so the integral would actually become $$\ 2 \large{\int}$$ $$u\ |u| \sin(|u|)\ du,\$$ which I'm not sure is correct. Basically the substitution $$\ x = u^2\$$ doesn't tell us if $$u=\sqrt{x}\$$ or $$\ u=-\sqrt{x}.\$$ Using the substitution $$\ u=\sqrt{x}\$$ leaves no room for ambiguity.

• I am very much concise :-( +1. – Sebastiano Jan 21 at 22:08
• Your answer was first and is more concise, but that doesn't automatically make it the best answer... Let the OP decide... – Adam Rubinson Jan 21 at 22:10
• Naturally but I am not interesting to have the upvotes :-) but upvoted the other users. Thus I am happy. – Sebastiano Jan 21 at 22:12
• I am selfish and want the upvotes. At least you can say I am honest about it though ;) – Adam Rubinson Jan 21 at 22:17
• Ha thanks! Much appreciated – Adam Rubinson Jan 21 at 22:25

The mistake is $$v'=\sin\sqrt{x}\,dx \to v=\int \sin\sqrt{x}\,dx\neq -\cos\sqrt{x}+k, \quad k\in \Bbb R$$

$$\int \sin\sqrt{x}\,dx=2\left(-\sqrt{x}\cos \left(\sqrt{x}\right)+\sin \left(\sqrt{x}\right)\right)+k',\quad k'\in \Bbb R$$

After the substitution $$u = \sqrt{x}$$, the integral becomes $$2\int u^2 \sin(u) \ du$$ as Adam Rubinson has said. We can proceed using tabular integration:

$$\begin{array}{c|c} u^2 & \sin(u) \\ \hline 2u & -\cos(u) \\ \hline \ 2 & -\sin(u) \\ \hline \ 0 & \cos(u)\end{array}$$

where we differentiate on the left and integrate on the right, and multiply the terms diagonally (the signs alternate).

Hence $$2\int u^2 \sin(u) \ du = 2 \left(-u^2 \cos(u) - -2u \sin(u) + 2\cos(u) \right) + C$$, which is:

$$2 \left( -x \cos( \sqrt x )+2\sqrt x \sin(\sqrt x) + 2\cos(\sqrt x)\right)+C = 4 \sqrt{x} \sin(\sqrt x) - 2(x-2)\cos (\sqrt x) +C$$