int {{e^{2x}}left( {frac{{sin 4x - 2}}{{1 - c | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) ધારો કે $I = \int {{e^{2x}}\left( {\frac{{\sin 4x - 2}}{{1 - \cos 4x}}} \right)} \,dx$. નિત્યસમ $\sin 4x = 2 \sin 2x \cos 2x$ અને $1 - \cos 4x = 2

Vedclass · Accepted Answer

$\frac{1}{2}{e^{2x}}\cot 2x + c$

Vedclass · Answer

(A) ધારો કે $I = \int {{e^{2x}}\left( {\frac{{\sin 4x - 2}}{{1 - \cos 4x}}} \right)} \,dx$. નિત્યસમ $\sin 4x = 2 \sin 2x \cos 2x$ અને $1 - \cos 4x = 2 \sin^2 2x$ નો ઉપયોગ કરતા: $I = \int {{e^{2x}}\left( {\frac{{2 \sin 2x \cos 2x - 2}}{{2 \sin^2 2x}}} \right)} \,dx$ $I = \int {{e^{2x}}\left( {\frac{{\sin 2x \cos 2x - 1}}{{\sin^2 2x}}} \right)} \,dx$ $I = \int {{e^{2x}}\left( {\cot 2x - \csc^2 2x} \right)} \,dx$ ધારો કે $f(x) = \cot 2x$,તો $f'(x) = -2 \csc^2 2x$. ખંડશઃ સંકલનનો ઉપયોગ કરતા: $\int e^{2x} \cot 2x \, dx - \int e^{2x} \csc^2 2x \, dx$. $\int e^{2x} \cot 2x \, dx = \frac{1}{2} e^{2x} \cot 2x - \int \frac{1}{2} e^{2x} (-2 \csc^2 2x) \, dx = \frac{1}{2} e^{2x} \cot 2x + \int e^{2x} \csc^2 2x \, dx$. આ કિંમત મૂકતા: $I = \left( \frac{1}{2} e^{2x} \cot 2x + \int e^{2x} \csc^2 2x \, dx \right) - \int e^{2x} \csc^2 2x \, dx + c$ $I = \frac{1}{2} e^{2x} \cot 2x + c$.

$\int {{e^{2x}}\left( {\frac{{\sin 4x - 2}}{{1 - \cos 4x}}} \right)\;dx = } $

Similar Questions

$\int \frac{e^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x, x>0=$

$\int \frac{x+100}{(x+101)^2} e^x \, dx = $ . . . . . . $+ C$.

$\int_0^1 \frac{e^x(x - 1)}{(x + 1)^3} \, dx = $

$\int {{e^x} \left[ \frac{1 + x \log x}{x} \right] \, dx} = $

$\int \left[ \frac{\log x - 1}{1 + (\log x)^2} \right]^2 dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module