Why can't I solve the following integral $\int \frac{\cos x}{5\sin x -3}\mathrm dx$? I tried using U-substitution to solve this but it didn't work. I set $u = \sin(x)$ and $\mathrm du = \cos(x)\mathrm dx$. I get $\int  \frac{\mathrm du}{5u-3}$. I get my final answer as $\ln(5\sin(x)-3)$. To be clear, I understand that setting $u = 5\sin(x)-3$ is another way of solving it. I am trying to understand why is it that U-substitution isn't working in this case. I am new integration to apologies for any mistakes.
 A: So from your comment to the post, it seems your issue stems from finding an antiderivative of
$$f(u)=\frac{1}{5u-3}.$$
For simplicity we suppose $u\in[-1,1]$, as that's the situation in your problems. An antiderivative of this would be
$$F(u)=\frac{1}{5}\ln(5u-3),$$
because if we differentiate it we get
$$F'(u)=\frac{1}{5}\cdot 5\cdot\frac{1}{5u-3}=\frac{1}{5u-3}=f(u),$$
where the factor of $5$ is the derivative of $5u-3$, i.e. the inner derivative we also have to respect. What you have to remember is that the chain rule gives rise to this factor of $5$ appearing when we differentiate, and so to compensate for that we need a factor of $\frac{1}{5}$ in the antiderivative. This can also be seen by making, for example, the substitution $t=5u-3$, $\mathrm{d}t=5~\mathrm{d}u$ in the indefinite integral
$$\int\frac{\mathrm{d}u}{5u-3}=\frac{1}{5}\int\frac{\mathrm{d}t}{t},$$
or simply by rewriting
$$\int\frac{\mathrm{d}u}{5u-3}=\frac{1}{5}\int\frac{\mathrm{d}u}{u-\frac{3}{5}}=\frac{1}{5}\ln\left\lvert u-\frac{3}{5}\right\rvert+C=\frac{1}{5}\ln\lvert 5u-3\rvert+\frac{1}{5}\ln \frac{1}{5}+C=\frac{1}{5}\ln\lvert 5u-3\rvert+D.$$
A: Just as a $u$ sub of
$u = 5\sin(x) - 3$ will send you looking for a function whose derivative is
[scalar] $~\times \dfrac{du}{u}$,
a $u$ sub of
$u = 5\sin(x)$ will send you looking for a function whose derivative is
[scalar] $~\times \dfrac{du}{u - 3}.$
