Q119 / 180 · KCET 2018 · Mathematics
\( \int e^{\sin x} \cdot\left(\frac{\sin x+1}{\sec x}\right) d x \) is equal to
- Correct answerA\( \sin x \cdot e^{\sin x}+C \)
- B\( \cos x \cdot e^{\sin x}+C \)
- C\( e^{\sin x}+C \)
- D\( e^{\sin x}(\sin x+1)+C \)
Solution
Given that, \( \int e^{\sin x}\left(\frac{\sin x+1}{\sec x}\right) d x \)
\[ =\int e^{\sin x}(\sin x \cos x+\cos x) d x \]
We know that,
\[ \begin{array}{l} \int e^{g(x)}\left(g^{\prime}(x) f(x)+f^{\prime}(x)\right) d x=e^{g(x)} f(x)+c \\ \text { So, } \int e^{\sin x}\left(\frac{\sin x+1}{\sec x}\right) d x=e^{\sin x} \sin x+c \end{array} \]
\[ =\int e^{\sin x}(\sin x \cos x+\cos x) d x \]
We know that,
\[ \begin{array}{l} \int e^{g(x)}\left(g^{\prime}(x) f(x)+f^{\prime}(x)\right) d x=e^{g(x)} f(x)+c \\ \text { So, } \int e^{\sin x}\left(\frac{\sin x+1}{\sec x}\right) d x=e^{\sin x} \sin x+c \end{array} \]
Hence, the correct answer is option: 'A'.