KCET · Maths · Probability
\( \int \frac{\sin ^{2} x}{1+\cos x} d x \)
- A \( x+\sin x+C \)
- B \( x-\sin x+C \)
- C \( \sin x+C \)
- D \( \cos x+C \)
Answer & Solution
Correct Answer
(B) \( x-\sin x+C \)
Step-by-step Solution
Detailed explanation
Given that, \( I=\int \frac{\sin ^{2} x}{1+\cos x} d x \)
\( =\int\left(\frac{1-\cos ^{2} x}{1+\cos x}\right) d x \)
\( =\int \frac{(1+\cos x)(1-\cos x)}{(1+\cos x)} d x \)
\( =\int(1-\cos x) d x=x-\sin x+c \)
\( =\int\left(\frac{1-\cos ^{2} x}{1+\cos x}\right) d x \)
\( =\int \frac{(1+\cos x)(1-\cos x)}{(1+\cos x)} d x \)
\( =\int(1-\cos x) d x=x-\sin x+c \)
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