int e^{2x+3} sin 6x , dx = | vedclass

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(A) ધારો કે $I = \int e^{2x+3} \sin 6x \, dx$. ખંડશઃ સંકલન (Integration by parts) $\int u \, dv = uv - \int v \, du$ નો ઉપયોગ કરતા,$u = \sin 6x$ અને $

Vedclass · Accepted Answer

$\frac{e^{2x+3}}{40}(2 \sin 6x - 6 \cos 6x) + C$

Vedclass · Answer

(A) ધારો કે $I = \int e^{2x+3} \sin 6x \, dx$. ખંડશઃ સંકલન (Integration by parts) $\int u \, dv = uv - \int v \, du$ નો ઉપયોગ કરતા,$u = \sin 6x$ અને $dv = e^{2x+3} \, dx$ લો. તેથી $du = 6 \cos 6x \, dx$ અને $v = \frac{e^{2x+3}}{2}$ મળે. $I = \frac{e^{2x+3}}{2} \sin 6x - \int \frac{e^{2x+3}}{2} \cdot 6 \cos 6x \, dx = \frac{e^{2x+3}}{2} \sin 6x - 3 \int e^{2x+3} \cos 6x \, dx$. ફરીથી $\int e^{2x+3} \cos 6x \, dx$ માટે ખંડશઃ સંકલન કરતા: $u = \cos 6x, dv = e^{2x+3} \, dx \implies du = -6 \sin 6x \, dx, v = \frac{e^{2x+3}}{2}$. $I = \frac{e^{2x+3}}{2} \sin 6x - 3 \left[ \frac{e^{2x+3}}{2} \cos 6x - \int \frac{e^{2x+3}}{2} (-6 \sin 6x) \, dx \right]$. $I = \frac{e^{2x+3}}{2} \sin 6x - \frac{3}{2} e^{2x+3} \cos 6x - 9 \int e^{2x+3} \sin 6x \, dx$. $I = \frac{e^{2x+3}}{2} \sin 6x - \frac{3}{2} e^{2x+3} \cos 6x - 9I$. $10I = \frac{e^{2x+3}}{2} (\sin 6x - 3 \cos 6x) + C$. $I = \frac{e^{2x+3}}{20} (\sin 6x - 3 \cos 6x) + C = \frac{e^{2x+3}}{40} (2 \sin 6x - 6 \cos 6x) + C$.

$\int e^{2x+3} \sin 6x \, dx =$

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