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