Question

${\displaystyle \frac{dy}{dx}+y \tan x = x^n \cos x, n\ne -1}$

Answer 1

We have,

${\displaystyle \frac{dy}{dx}+y\tan x=x^n\cos x}$

$$\text{Comparing with }\frac{dy}{dx}+Py=Q,\text{ we get}$$

$$P=\tan x$$

$$Q=x^n\cos x$$

Now,

$$I.F.=e^{\int \tan xdx}$$

$$=e^{\log \left(secx\right)}$$

$$=\sec x$$

So, the solution is given by

$$y\times I.F.=\int Q\times I.F.dx+C$$

$$\Rightarrow y\sec x=\int x^n\cos x\sec xdx+C$$

$$\Rightarrow y\sec x=\int x^{n}dx+C$$

$$\Rightarrow y\sec x=\frac{x^{n+1}}{n+1}+C$$