Question

$\int e^{\tan x}(\sec ^2 x+\sec ^3 x \sin x) d x=$

• A. $\tan x \cdot e^{\tan x}+c$
• B. $(1+\tan x) e^{\tan x}+c$
• C. $\sec x \cdot e^{\tan x}+c$
• D. $e^{\tan x}+\tan x+c$

Hint:

If you see $e^{f(x)}$ and a term involving $f'(x)$, substitution is usually the fastest path to the solution.

Answer 1

The Correct Option is A

Step 1: Tidy the integrand.
Pull out $\sec^2 x$: $\sec^2 x + \sec^3 x\sin x = \sec^2 x(1 + \sec x\sin x)$. Since $\sec x\sin x = \tan x$, the bracket is $1+\tan x$.

Step 2: Substitute.
Let $u = \tan x$, so $du = \sec^2 x\,dx$. The integral becomes \[ \int e^u (1+u)\,du. \]

Step 3: Recognise the form.
This is the classic $\int e^u(f(u)+f'(u))\,du = e^u f(u)$ with $f(u)=u$ and $f'(u)=1$. So the answer is $u\,e^u + c$.

Step 4: Go back to $x$.
\[ \boxed{\tan x\cdot e^{\tan x} + c} \]