Question

\[\frac{dy}{dx} = \sin^3 x \cos^2 x + x e^x\]

Answer 1

We have,

\[\frac{dy}{dx} = \sin^3 x \cos^2 x + x e^x \]

\[ \Rightarrow dy = \left( \sin^3 x \cos^2 x + x e^x \right)dx\]

Integrating both sides, we get

\[\int dy = \int\left( \sin^3 x \cos^2 x + x e^x \right)dx\]

\[ \Rightarrow y = \int \sin^3 x \cos^2 x dx + \int x e^x dx \]

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

Here,

\[ I_1 = \int \sin^3 x \cos^2 x dx\]

\[ I_2 = \int x e^x dx\]

Now,

\[ I_1 = \int \sin^3 x \cos^2 x dx\]

\[ = \int\left( 1 - \cos^2 x \right) \cos^2 x \sin x dx\]

\[\text{Putting }t = \cos x,\text{ we get}\]

\[dt = - \sin x dx\]

\[ \therefore I_1 = - \int t^2 \left( 1 - t^2 \right)dt\]

\[ = \int - t^2 + t^4 dt\]

\[ = - \frac{t^3}{3} + \frac{t^5}{5} + C_1 \]

\[ = \frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C_1 \]

\[ I_2 = \int x e^x dx\]

\[ = x\int e^x dx - \int\left( \frac{d}{dx}\left( x \right)\int e^x dx \right)dx\]

\[ = x e^x - e^x + C_2 \]

\[ = \left( x - 1 \right) e^x + C_2 \]

\[\text{Putting the value of }I_1\text{ and }I_2\text{ in (1), we get}\]

\[y = \frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C_1 + \left( x - 1 \right) e^x + C_2 \]
\[y = \frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + \left( x - 1 \right) e^x + C,\text{ where }C = C_1 + C_2\]