Question

The equation of a curve passing through origin is given by $y=\int x^{3}cos{x}^{4}dx$. If the equation of the curve is written in the form x = g(y), then

• A. $g\left(y\right)=\sqrt[3]{3si{n}^{-1}\left(4y\right)}$
• B. $g\left(y\right)=\sqrt{si{n}^{-1}\left(4y\right)}$
• C. $g\left(y\right)=\sqrt[4]{4si{n}^{-1}\left(4y\right)}$
• D. None of these

Answer 1

The correct option is C $g\left(y\right)=\sqrt[4]{4si{n}^{-1}\left(4y\right)}$

We have, $y=\int x^{3}cos{x}^{4}dx=\frac{1}{4}\int costdt[Put{x}^4=t\Rightarrow x^{3}dx=\frac{1}{4}dt]=\frac{1}{4}sint+C=\frac{1}{4}sin{x}^4+C$
Since the curve passes through (0, 0) ,therefore C=0
$\therefore y=\frac{1}{4}sin{x}^4\Rightarrow x^4=si{n}^{-1}\left(4y\right)\Rightarrow x=\sqrt[4]{4si{n}^{-1}\left(4y\right)}$