Calculus 2: Inverse Trig Functions

Question

I'm stuck on the following problem:

I'm looking at the solution provided by the teacher but I think it might be wrong? Please let me know if the solution is wrong ... and if so where...

I personally think it should be $$\frac{7}{5}\sin^{-1}(\frac{7x}{5})$$ ....

Answer 1

Hint:

Regarding your statement of

I personally think it should be 7/5(sin^-1(7x/5) ....

note that by the chain rule, you have

$$\begin{equation}\begin{aligned} \frac{d\left(\sin^{-1}\left(\frac{7x}{5}\right)\right)}{dx} & = \frac{d\left(\sin^{-1}\left(\frac{7x}{5}\right)\right)}{d\left(\frac{7x}{5}\right)}\frac{d{\left(\frac{7x}{5}\right)}}{dx} \\ & = \left(\frac{1}{\sqrt{1-\left(\frac{7x}{5}\right)^2}}\right)\left(\frac{7}{5}\right) \\ & = \frac{7}{5\sqrt{1-\left(\frac{7x}{5}\right)^2}} \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Answer 2

The answer is correct except that a minus sign is missing here: $$u=25-49x^2 \implies du=\color{red}{-98}dx$$

For the arcsin the integral is correctly evaluated: $$I=\int \dfrac {d\left(\dfrac 75 x\right)}{\sqrt {1-\left(\dfrac 75 x \right)^2}}=\arcsin \left(\dfrac 75 x\right)$$ Just substitute $u=\dfrac 75 x \implies du=\dfrac 75 dx$ $$I=\int \dfrac {du}{\sqrt {1-u^2}}=\arcsin u=\arcsin\left(\dfrac 75 x\right)$$