int frac{e^{cot x}}{sin^2 x} (2 log csc x + s | vedclass

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Question

(B) Let $I = \int \frac{e^{\cot x}}{\sin^2 x} (2 \log \csc x + \sin 2 x) dx$. Substitute $t = \cot x$,then $dt = -\csc^2 x dx$,which implies $\csc^2 x

Vedclass · Accepted Answer

$-2 e^{\cot x} \log (\csc x) + C$

Vedclass · Answer

(B) Let $I = \int \frac{e^{\cot x}}{\sin^2 x} (2 \log \csc x + \sin 2 x) dx$. Substitute $t = \cot x$,then $dt = -\csc^2 x dx$,which implies $\csc^2 x dx = -dt$. Also,$\sin 2x = 2 \sin x \cos x = \frac{2 \sin x \cos x}{\sin^2 x} \cdot \sin^2 x = 2 \cot x \sin^2 x$. Wait,let us simplify the integrand: $\frac{1}{\sin^2 x} = \csc^2 x$. So,$I = \int e^{\cot x} \csc^2 x (2 \log \csc x + 2 \sin x \cos x) dx$. Since $\sin 2x = 2 \sin x \cos x$,we have $\frac{\sin 2x}{\sin^2 x} = 2 \cot x$. Thus,$I = \int e^{\cot x} (2 \csc^2 x \log \csc x + 2 \cot x \csc^2 x) dx$. Let $u = \cot x$,then $du = -\csc^2 x dx$. Note that $\csc^2 x = 1 + \cot^2 x = 1 + u^2$. Also $\log \csc x = \log (1 + u^2)^{1/2} = \frac{1}{2} \log (1 + u^2)$. Substituting these into the integral: $I = -\int e^u (2 \cdot \frac{1}{2} \log (1 + u^2) + 2u) du = -\int e^u (\log (1 + u^2) + 2u) du$. This does not simplify directly to the standard form $\int e^u (f(u) + f'(u)) du$. Let us re-evaluate: $\frac{d}{dx} (e^{\cot x} \log \csc^2 x) = e^{\cot x} (-\csc^2 x) \log \csc^2 x + e^{\cot x} \frac{1}{\csc^2 x} (2 \csc x \cdot -\csc x \cot x) = -e^{\cot x} \csc^2 x \log \csc^2 x - 2 e^{\cot x} \cot x$. Comparing with the integrand,the correct form is $-2 e^{\cot x} \log \csc x + C$.

$\int \frac{e^{\cot x}}{\sin^2 x} (2 \log \csc x + \sin 2 x) dx =$

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