Finding an integrand when solution is known

Question

I am looking for an integrand $f(x)$ such that $$ \int^{t_2}_{t_1}f(b,a,x)dx=a \log(\cos(b(t_2-t_1))) $$ where $a, t_1,t_2$ and $b$ are constants and $a$ can be set to $1$ if needed. I know the special case when $t_1=0$, $a=-\frac{1}{b}$ the integrand is $f(x)=\tan(b x)$. But I don't know how to get the above output.

Answer 1

Partial derivatives are helpful when there are more than one variables. Suppose you have:

$$\int_{t_{1}}^{t_{2}}f\left(x\right)dx=F\left(t_{1},t_{2}\right)$$

Partial derivative w.r.t $t_1$, By Newton-Leibnitz differentiation, we have:

$$f\left(t_{1}\right)=F'\left(t_{1},t_{2}\right)$$

For your function, we get:

$$f\left(t_{1}\right)=-ab\tan\left(t_{1}-t_{2}\right)$$

If you integrate this with respect to $t_1$ with the same limits, you will get the same result. Otherwise, you cannot expect a two-variable function to be converted to a single variable

Answer 2

$$\begin{align}\int_{t_1}^{t_2}f(x) dx&=\int_{0}^{t_2-t_1} f(x+t_1)dx \\ &=G(t_2-t_1)-G(0)=a\ln\cos(b(t_2-t_1))\end{align}$$

Assume $t_2-t_1=c$.

$$f(c+t_1)=g(c)=\frac{d}{dc}(G(c)-G(0))=\frac{d}{dc}a\ln\cos(bc)=-ab\tan(bc)$$

Then $f(t_2)=-ab\tan(b(t_2-t_1))$.

If you wanted to find $F(x)$ that satisfies $F(t_2)-F(t_1)=a \log \cos b(t_2-t_1)$, note the following:

$$F(x)-F(0)=a\log\cos bx$$

Therefore, $F(x)=a\log\cos bx+F(0)$. This means $$a\log\cos bt_2-a\log\cos bt_1=a\log\cos b(t_2-t_1)$$ becomes an equation, not an identity. For this to hold, $\sin t_1=0$ or $\sin(t_2-t_1)=0$ must be. The function is undefined if it cannot be the solution to this equation.