MHT CET · Maths · Indefinite Integration
\(\int e^x\left(\frac{1+\sin x}{1+\cos x}\right) d x=\)
- A \(\mathrm{e}^{\mathrm{x}} \tan \frac{\mathrm{x}}{2}+\mathrm{c}\)
- B \(e^x \cot \frac{x}{2}+c\)
- C \(\mathrm{e}^{\mathrm{x}} \cos \frac{\mathrm{x}}{2}+\mathrm{c}\)
- D \(e^x \sin \frac{x}{2}+c\)
Answer & Solution
Correct Answer
(A) \(\mathrm{e}^{\mathrm{x}} \tan \frac{\mathrm{x}}{2}+\mathrm{c}\)
Step-by-step Solution
Detailed explanation
\(I=\int e^x\left(\frac{1+\sin x}{1+\cos x}\right) d x=\int e^x(\frac{1}{1+\cos x}+\) \(\frac{\sin x}{1+\cos x}) d x\)
\(=\int e^x\left(\frac{1}{2 \cos ^2 \frac{x}{2}}+\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}\right) d x=\int e^x\left(\frac{1}{2}\right)\) \(\left(\sec ^2 \frac{x}{2}+2 \tan \frac{x}{2}\right) d x\)
\(=\frac{1}{2} \int e^x\left[2 \tan \frac{x}{2}+\sec ^2 \frac{x}{2}\right] d x=\frac{1}{2} \cdot e^x(2)+\tan \left(\frac{x}{2}\right)\) \(+c=e^x \tan \frac{x}{2}+c\)
\(=\int e^x\left(\frac{1}{2 \cos ^2 \frac{x}{2}}+\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}\right) d x=\int e^x\left(\frac{1}{2}\right)\) \(\left(\sec ^2 \frac{x}{2}+2 \tan \frac{x}{2}\right) d x\)
\(=\frac{1}{2} \int e^x\left[2 \tan \frac{x}{2}+\sec ^2 \frac{x}{2}\right] d x=\frac{1}{2} \cdot e^x(2)+\tan \left(\frac{x}{2}\right)\) \(+c=e^x \tan \frac{x}{2}+c\)
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