Question

\[(\tan^2 x + 2\tan x + 5)\frac{dy}{dx} = 2(1+\tan x)\sec^2x\]

Answer 1

We have,

\[\left( \tan^2 x + 2 \tan x + 5 \right)\frac{dy}{dx} = 2\left( 1 + \tan x \right) \sec^2 x\]

\[ \Rightarrow dy = \frac{2\left( 1 + \tan x \right) \sec^2 x}{\left( \tan^2 x + 2 \tan x + 5 \right)} dx\]

Integrating both sides, we get

\[\int dy = \int\frac{2\left( 1 + \tan x \right) \sec^2 x}{\left( \tan^2 x + 2 \tan x + 5 \right)} dx . . . . . . . . \left( 1 \right)\]
\[\text{Putting }\tan^2 x + 2 \tan x + 5 = t\]

\[ \therefore \left( 2 \tan x se c^2 x + 2se c^2 x \right) dx = dt\]

\[ \Rightarrow 2\left( 1 + \tan x \right) \sec^2 x dx = dt\]

Therefore (1) becomes,

\[\int dy = \int\frac{1}{t} dt\]

\[ \Rightarrow y = \log \left| t \right| + C\]

\[ \Rightarrow y = \log \left| \tan^2 x + 2 \tan x + 5 \right| + C\]