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