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\(\int e^{x / 2}\left(\frac{2+\sin x}{1+\cos x}\right) d x=\)
Options:
Solution:
1965 Upvotes
Verified Answer
The correct answer is:
\(2 e^{x / 2} \tan \left(\frac{x}{2}\right)+c\)
\(\begin{aligned}
& \int e^{\frac{x}{2}}\left(\frac{2+\sin x}{1+\cos x}\right) d x \\
& \int e^{\frac{x}{2}}\left(\frac{2+\frac{2 \tan x / 2}{1+\tan ^2 x / 2}}{1+\frac{1-\tan ^2 x / 2}{1+\tan ^2 x / 2}}\right) d x \\
& \int e^{\frac{x}{2}}\left(\frac{2+2 \tan ^2 x / 2+2 \tan x / 2}{1+\tan ^2 x / 2+1-\tan ^2 x / 2}\right) d x
\end{aligned}\)
\(\begin{gathered}
\int e^{\frac{x}{2}} \cdot 2\left(\frac{1+\tan ^2 x / 2+\tan x / 2}{2}\right) d x \\
\int e^{\frac{x}{2}}\left(\sec ^2 \frac{x}{2}+\tan \frac{x}{2}\right) d x \\
\text {Put, } \frac{x}{2}=t \\
d x=2 d t \\
\int e^t\left(\sec ^2 t+\tan t\right) \cdot 20 d t \\
2 \cdot e^t \tan t+c \\
2 \cdot e^{\frac{x}{2}} \tan \frac{x}{2}+c
\end{gathered}\)
Hence, option (b) is correct.
& \int e^{\frac{x}{2}}\left(\frac{2+\sin x}{1+\cos x}\right) d x \\
& \int e^{\frac{x}{2}}\left(\frac{2+\frac{2 \tan x / 2}{1+\tan ^2 x / 2}}{1+\frac{1-\tan ^2 x / 2}{1+\tan ^2 x / 2}}\right) d x \\
& \int e^{\frac{x}{2}}\left(\frac{2+2 \tan ^2 x / 2+2 \tan x / 2}{1+\tan ^2 x / 2+1-\tan ^2 x / 2}\right) d x
\end{aligned}\)
\(\begin{gathered}
\int e^{\frac{x}{2}} \cdot 2\left(\frac{1+\tan ^2 x / 2+\tan x / 2}{2}\right) d x \\
\int e^{\frac{x}{2}}\left(\sec ^2 \frac{x}{2}+\tan \frac{x}{2}\right) d x \\
\text {Put, } \frac{x}{2}=t \\
d x=2 d t \\
\int e^t\left(\sec ^2 t+\tan t\right) \cdot 20 d t \\
2 \cdot e^t \tan t+c \\
2 \cdot e^{\frac{x}{2}} \tan \frac{x}{2}+c
\end{gathered}\)
Hence, option (b) is correct.
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