MHT CET · Maths · Definite Integration
The value of \(\int_{\pi / 4}^{\pi / 2} e^{x}(\log \sin x+\cot x) d x\) is
- A \(e^{\pi / 4} \log 2\)
- B \(-e^{\pi / 4} \log 2\)
- C \(\frac{1}{2} e^{\pi / 4} \log 2\)
- D \(-\frac{1}{2} e^{\pi / 4} \log 2\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2} e^{\pi / 4} \log 2\)
Step-by-step Solution
Detailed explanation
Let
\(I=\int_{\pi / 4}^{\pi / 2} e^{x}(\log \sin x +\cot x) d x \)
\( \Rightarrow I=\int_{\pi / 4}^{\pi / 2} e^{x} \log \sin x d x \) \( +\int_{\pi / 4}^{\pi / 2} e^{x} \cot x d x \)
\( =\int_{\pi / 4}^{\pi / 2} e^{x} \log \sin x d x+\left[e^{x} \log \sin x\right]_{\pi / 4}^{\pi / 2} \) \( -\int_{\pi / 4}^{\pi / 2} e^{x} \log \sin x d x \)
\( = e^{\pi / 2} \log \sin \frac{\pi}{2}-e^{\pi / 4} \log \sin \frac{\pi}{4} \)
\( = -e^{\pi / 4} \log \left(\frac{1}{\sqrt{2}}\right) \)
\( = \frac{1}{2} e^{\pi / 4} \log 2\)
\(I=\int_{\pi / 4}^{\pi / 2} e^{x}(\log \sin x +\cot x) d x \)
\( \Rightarrow I=\int_{\pi / 4}^{\pi / 2} e^{x} \log \sin x d x \) \( +\int_{\pi / 4}^{\pi / 2} e^{x} \cot x d x \)
\( =\int_{\pi / 4}^{\pi / 2} e^{x} \log \sin x d x+\left[e^{x} \log \sin x\right]_{\pi / 4}^{\pi / 2} \) \( -\int_{\pi / 4}^{\pi / 2} e^{x} \log \sin x d x \)
\( = e^{\pi / 2} \log \sin \frac{\pi}{2}-e^{\pi / 4} \log \sin \frac{\pi}{4} \)
\( = -e^{\pi / 4} \log \left(\frac{1}{\sqrt{2}}\right) \)
\( = \frac{1}{2} e^{\pi / 4} \log 2\)
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