Integral Calculus
86352
\(\int e^{-x} \operatorname{cosec} x(1+\cot x) d x\) is equal to
1 \(e^{-x} \operatorname{cosec} x+C\)
2 \(-\mathrm{e}^{-\mathrm{x}} \operatorname{cosec} \mathrm{x}+\mathrm{C}\)
3 \(-e^{-x}(\operatorname{cosec} x+\cot x)+C\)
4 \(-\mathrm{e}^{-\mathrm{x}}(\operatorname{cosec} \mathrm{x}-\tan \mathrm{x})+C\)
5 \(-\mathrm{e}^{-\mathrm{x}} \sec \mathrm{x}+C\)
Explanation:
(B) : Given,
\(I=\int e^{-x} \operatorname{cosec} x(1+\cot x) d x\)
Let, \(\quad e^{-x} \operatorname{cosec} x=t\)
\(\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{-\mathrm{x}} \operatorname{cosec} \mathrm{x}\right)=\frac{\mathrm{dt}}{\mathrm{dx}}\)
\(-e^{-x} \operatorname{cosec} x-e^{-x} x \operatorname{cosec} x \cdot \cot x=\frac{d t}{d x}\)
\(-e^{-x} \operatorname{cosec} x(1+\cot x)=\frac{d t}{d x}\)
\(d x=\frac{1}{-e^{x} \operatorname{cosec}(1+\cot x)} d t\)
\(\text { Now, }\)
\(I=\int e^{-x} \operatorname{cosec} x(1+\cot x) \cdot \frac{1}{-e^{-x} \operatorname{cosec} x(1+\cot x)} d t\)
\(I=\int-d t \Rightarrow I=-t+C\)
\(I=-e^{-x} \operatorname{cosec} x+C\)
