KCET · Maths · Indefinite Integration
The value of \(\int e^{x}\left[\frac{1+\sin x}{1+\cos x}\right] d x\) is equal to
- A \(e^{x} \tan \frac{x}{2}+C\)
- B \(e^{x} \tan x+C\)
- C \(e^{x}(1+\cos x)+C\)
- D \(e^{x}(1+\sin x)+C\)
Answer & Solution
Correct Answer
(A) \(e^{x} \tan \frac{x}{2}+C\)
Step-by-step Solution
Detailed explanation
Let \(I=\int e^{x}\left(\frac{1+\sin x}{1+\cos x}\right) d x\)
\(=\int e^{x}\left(\frac{1+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\right) d x\)
\(=\int e^{x}\left[\frac{\sec ^{2} \frac{x}{2}}{2}+\tan \frac{x}{2}\right] d x\)
Let \(f(x)=\tan \frac{x}{2}\)
\(\therefore \quad f^{\prime}(x)=\frac{\sec ^{2} \frac{x}{2}}{2}\)
Using the formula, \(\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x\)
\(\begin{aligned}
&=e^{x} f(x)+C \\
\Rightarrow \quad I &=e^{x} \times \tan \frac{x}{2}+C
\end{aligned}\)
\(=\int e^{x}\left(\frac{1+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\right) d x\)
\(=\int e^{x}\left[\frac{\sec ^{2} \frac{x}{2}}{2}+\tan \frac{x}{2}\right] d x\)
Let \(f(x)=\tan \frac{x}{2}\)
\(\therefore \quad f^{\prime}(x)=\frac{\sec ^{2} \frac{x}{2}}{2}\)
Using the formula, \(\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x\)
\(\begin{aligned}
&=e^{x} f(x)+C \\
\Rightarrow \quad I &=e^{x} \times \tan \frac{x}{2}+C
\end{aligned}\)
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