Q111 / 180 · COMEDK 2018 · Mathematics
$\int e^{x}\left\{\frac{1+\sin x \cos x}{\cos ^{2} x}\right\} d x=$
- A$e^{x} \cos x+c$
- B$e^{x} \sec x \tan x+c$
- Correct answerC$e^{x} \tan x+c$
- D$e^{x} \cos ^{2} x-1+c$
Solution
We have,
$$
\begin{aligned}
&\int e^{x}\left\{\frac{1+\sin x \cos x}{\cos ^{2} x}\right\} d x \\
&=\int e^{x}\left\{\frac{1}{\cos ^{2} x}+\frac{\sin x \cos x}{\cos ^{2} x}\right\} d x \\
&=\int e^{x}\left(\sec ^{2}+\tan x\right) d x=\int e^{x}\left\{f^{\prime}(x)+f(x)\right\} d x \\
&{\left[\text { where } f(x)=\tan x \text { and } f^{\prime}(x)=\sec ^{2} x\right]} \\
&=e^{x} f(x)+c=e^{x} \tan x+c
\end{aligned}
$$
Hence, the correct answer is option: 'C'.