# I tried to solve $\int t \sin (t^2) dt$ using partial integration. [duplicate]

I tried to solve $$\int t \sin (t^2) dt$$ using partial integration. The result is $$\frac{-t \cos{(t^2)}}{2} + \frac{\sin{(t^2)}}{4} + C$$.

Then I tried to differente that result. Instead of getting $$t \sin (t^2)$$, I got $$t^2 \sin{(t^2)}-\frac{\cos{(t^2)}}{2}+\frac{t\cos{(t^2)}}{2}$$.

What's wrong? (I know I can use substitution, just curious to use partial integration.)

Edit: Here is how I get the result: Using the formula $$\int uv' dt = uv-\int u'vdt$$.
$$u = t \rightarrow u \frac{d}{dt} = t \frac{d}{dt} \rightarrow u \frac{d}{dt} = 1$$, and $$v \frac{d}{dt} = \sin{(t^2)} \rightarrow \int {v \frac{d}{dt}} dt = \int{\sin{(t^2)}} dt \rightarrow v = \frac{-\cos {(t^2)}}{2}+C_1$$, so \begin{align} uv-\int{u'v}dt = \frac{-t \cos{(t^2)}}{2} + tC_1 - \int{\frac{-\cos{(t^2)}}{2}}dt - \int{C_1}dt = \frac{-t \cos{(t^2)} }{2} + tC_1 + \frac{ \sin{(t^2)}}{4} - tC_1 + C_2 = \frac{-t \cos{(t^2)} }{2} + \frac{ \sin{(t^2)}}{4} + C_2. \end{align}

• Using substitution, as you say, the result is $\int{t\sin{(t^{2})}\,\mathrm{d}t} = -\cos{(t^{2})}/2$, so that should be the answer you get, right? Can you elaborate on how you obtained your result? Commented Jan 20, 2023 at 13:08
• Commented Jan 21, 2023 at 8:01
• Use Approach0. Commented Jan 21, 2023 at 8:07

I’m guessing that you integrated $$\sin t^2$$ to $$-\frac12\cos t^2$$ and $$\cos t^2$$ to $$\frac12\sin t^2$$, as that would yield the result you give. That’s wrong, as you can see if you differentiate the results.
\begin{align} \int \underbrace{t}_u\underbrace{\sin\left(t^2\right)}_{v'}\text{d}t = \underbrace{t}_u \cdot \underbrace{\int \sin\left(t^2\right)\text{d}t}_v - \int \left(\underbrace{1}_{u'} \cdot \underbrace{\int \sin\left(t^2\right)\text{d}t}_v \right)\text{d}t \tag{1}\label{1} \end{align} Your mistake lies in the integration of $$v'$$. Following your statement would mean: $$v \frac{\text{d}}{\text{d}t} = - \frac{1}{2}\cos\left(t^2\right) \frac{\text{d}}{\text{d}t} \color{red}{ = \sin\left(t^2\right) = v'}$$ The correct derivative of $$-\frac{1}{2}\cos\left(t^2\right)$$ would be $$t \sin\left(t^2\right)$$ (which is what you started with so you accidentally found the correct integral for the initial function).
The correct integral of $$v' = \sin\left(t^2\right)$$ would be: \begin{align} v = \int \sin\left(t^2\right)\text{d}t &= \frac{1}{4} \left(1 + i\right)\left(-i\operatorname{erf}\left(\frac{1}{2}\left(1-i\right)t\sqrt{\pi}\sqrt{\frac{2}{/pi}}\right)+\operatorname{erf}\left(\frac{1}{2}\left(1+i\right)t\sqrt{\pi}\sqrt{\frac{2}{\pi}}\right)\right)\sqrt{\frac{\pi}{2}} \\ &= \left(\frac{1}{4}+\frac{i}{4}\right)\sqrt{\frac{\pi}{2}}\left(\operatorname{erf}\left(\frac{\left(1+i\right)t}{\sqrt{2}}\right)-\operatorname{erfi}\left(\frac{\left(1+i\right)t}{\sqrt{2}}\right)\right) \end{align} As you can see, this wont integrate nicely and you will need the Error function. Putting this into $$\eqref{1}$$ and integrating it a second time will eventually get you to the same solution but substitution is obviously the better choice.