∫ x^3√(7-x^2)\ dx is equal to(where C is an arbitrary constant)

The correct answer is Option (3) → \( -\displaystyle\frac{7}{3}(7-x^2)^{3/2}+\displaystyle\frac{1}{5}(7-x^2)^{5/2}+C \)

Since \( x^3=x^2\cdot x \), the substitution \( t=7-x^2 \) works, with \( x^2=7-t \) and \( x\,dx=-\frac{1}{2}dt \).

    \( \displaystyle \int x^3\sqrt{7-x^2}\,dx=\int x^2\sqrt{7-x^2}\cdot x\,dx \)

    \( \displaystyle =-\frac{1}{2}\int(7-t)\,t^{1/2}\,dt \)

    \( \displaystyle =-\frac{1}{2}\int\left(7t^{1/2}-t^{3/2}\right)dt \)

    \( \displaystyle =-\frac{1}{2}\left[\frac{14}{3}t^{3/2}-\frac{2}{5}t^{5/2}\right]+C \)

    \( \displaystyle =-\frac{7}{3}t^{3/2}+\frac{1}{5}t^{5/2}+C \)

Putting back \( t=7-x^2 \):

    \( \displaystyle \int x^3\sqrt{7-x^2}\,dx=-\frac{7}{3}(7-x^2)^{3/2}+\frac{1}{5}(7-x^2)^{5/2}+C \)

Shortcut: Differentiate Option (3) instead, with \( t=7-x^2 \) and \( \frac{dt}{dx}=-2x \):

    \( \displaystyle \frac{d}{dx}\left[-\frac{7}{3}t^{3/2}+\frac{1}{5}t^{5/2}\right] \)

    \( \displaystyle =-\frac{7}{2}t^{1/2}(-2x)+\frac{1}{2}t^{3/2}(-2x) \)

    \( \displaystyle =x\sqrt{t}\,(7-t) \)

Since \( 7-t=x^2 \), this is \( x^3\sqrt{7-x^2} \), the integrand.