Question

\[\frac{dy}{dx} = \sin^3 x \cos^4 x + x\sqrt{x + 1}\]

Answer 1

We have,

\[\frac{dy}{dx} = \sin^3 x \cos^4 x + x\sqrt{x + 1}\]

\[ \Rightarrow dy = \left( \sin^3 x \cos^4 x + x\sqrt{x + 1} \right)dx\]

Integrating both sides, we get

\[\int dy = \int\left( \sin^3 x \cos^4 x + x\sqrt{x + 1} \right)dx\]

\[ \Rightarrow y = \int \sin^3 x \cos^4 x dx + \int x\sqrt{x + 1}dx \]

\[ \Rightarrow y = I_1 + I_2 . . . . . \left( 1 \right) \]

Here,

\[ I_1 = \int \sin^3 x \cos^4 x dx\]

\[ I_1 = \int x\sqrt{x + 1}dx\]
Now,
\[ I_1 = \int \sin^3 x \cos^4 x dx\]
\[ = \int\left( 1 - \cos^2 x \right) \cos^4 x \sin x dx\]
\[\text{Putting }t = \cos x,\text{ we get}\]
\[dt = - \sin x dx\]
\[ \therefore I_1 = - \int t^4 \left( 1 - t^2 \right)dt\]
\[ = \int\left( t^6 - t^4 \right)dt\]
\[ = \frac{t^7}{7} - \frac{t^5}{5} + C_1 \]
\[ = \frac{\cos^7 x}{7} - \frac{\cos^5 x}{5} + C_1 \]
\[ I_2 = \int x\sqrt{x + 1}dx\]
\[\text{Putting }t^2 = x + 1,\text{ we get}\]
\[2t dt = dx\]
\[ \therefore I_2 = 2\int\left( t^2 - 1 \right) t^2 dt\]
\[ = 2\int\left( t^4 - t^2 \right) dt\]
\[ = \frac{2 t^5}{5} - \frac{2 t^3}{3} + C_2 \]
\[ = \frac{2 \left( x + 1 \right)^\frac{5}{2}}{5} - \frac{2 \left( x + 1 \right)^\frac{3}{2}}{3} + C_2 \]
\[\text{Putting the value of }I_1\text{ and }I_2\text{ in (1), we get}\]
\[y = \frac{\cos^7 x}{7} - \frac{\cos^5 x}{5} + C_1 + \frac{2 \left( x + 1 \right)^\frac{5}{2}}{5} - \frac{2 \left( x + 1 \right)^\frac{3}{2}}{3} + C_2 \]
\[y = \frac{\cos^7 x}{7} - \frac{\cos^5 x}{5} + \frac{2 \left( x + 1 \right)^\frac{5}{2}}{5} - \frac{2 \left( x + 1 \right)^\frac{3}{2}}{3} + C .............\left[\because C = C_1 + C_2 \right]\]
\[\text{Hence, }y = \frac{\cos^7 x}{7} - \frac{\cos^5 x}{5} + \frac{2 \left( x + 1 \right)^\frac{5}{2}}{5} - \frac{2 \left( x + 1 \right)^\frac{3}{2}}{3} + C\text{ is the solution of the given differential equation.}\]