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