int {{e^{ - x}}{{csc }^2}(2{e^{ - x}} + 5)} , | vedclass

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Question

(A) Let $I = \int {{e^{ - x}}{{\csc }^2}(2{e^{ - x}} + 5)} \,dx$. Substitute $t = 2{e^{ - x}} + 5$. Then,$dt = -2{e^{ - x}}dx$,which implies ${e^{ - x

Vedclass · Accepted Answer

$\frac{1}{2}\cot (2{e^{ - x}} + 5) + c$

Vedclass · Answer

(A) Let $I = \int {{e^{ - x}}{{\csc }^2}(2{e^{ - x}} + 5)} \,dx$. Substitute $t = 2{e^{ - x}} + 5$. Then,$dt = -2{e^{ - x}}dx$,which implies ${e^{ - x}}dx = -\frac{1}{2}dt$. Substituting these into the integral: $I = \int {\csc^2(t) \cdot (-\frac{1}{2}dt)} = -\frac{1}{2} \int \csc^2(t) dt$. Since $\int \csc^2(t) dt = -\cot(t) + c$,we get: $I = -\frac{1}{2} (-\cot(t)) + c = \frac{1}{2} \cot(t) + c$. Substituting back $t = 2{e^{ - x}} + 5$,we obtain: $I = \frac{1}{2} \cot(2{e^{ - x}} + 5) + c$.

$\int {{e^{ - x}}{{\csc }^2}(2{e^{ - x}} + 5)} \,dx = $

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