KCET · Maths · Indefinite Integration
\( \int e^{\sin x} \cdot\left(\frac{\sin x+1}{\sec x}\right) d x \) is equal to
- A \( \sin x \cdot e^{\sin x}+C \)
- B \( \cos x \cdot e^{\sin x}+C \)
- C \( e^{\sin x}+C \)
- D \( e^{\sin x}(\sin x+1)+C \)
Answer & Solution
Correct Answer
(A) \( \sin x \cdot e^{\sin x}+C \)
Step-by-step Solution
Detailed explanation
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,
\[
\begin{array}{l}
\int e^{g(x)}\left(g^{\prime}(x) f(x)+f^{\prime}(x)\right) d x=e^{g(x)} f(x)+c \\
\text { So, } \int e^{\sin x}\left(\frac{\sin x+1}{\sec x}\right) d x=e^{\sin x} \sin x+c
\end{array}
\]
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