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