int frac{x + sin x}{1 + cos x} dx = | vedclass

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

Question

(D) ધારો કે $I = \int \frac{x + \sin x}{1 + \cos x} dx$ $\Rightarrow I = \int \frac{x}{1 + \cos x} dx + \int \frac{\sin x}{1 + \cos x} dx$ ધારો કે $I

Vedclass · Accepted Answer

$x 	an \frac{x}{2} + c$

Vedclass · Answer

(D) ધારો કે $I = \int \frac{x + \sin x}{1 + \cos x} dx$ $\Rightarrow I = \int \frac{x}{1 + \cos x} dx + \int \frac{\sin x}{1 + \cos x} dx$ ધારો કે $I = I_{1} + I_{2}$ હવે,$I_{1} = \int \frac{x}{1 + \cos x} dx = \int \frac{x \ dx}{2 \cos^{2} \frac{x}{2}}$ $\left\{ \because 1 + \cos 	heta = 2 \cos^{2} \frac{	heta}{2} ight\}$ $\Rightarrow I_{1} = \int \frac{x}{2} \sec^{2} \frac{x}{2} dx$ ખંડશઃ સંકલનનો ઉપયોગ કરતા: $\int u v \ dx = u \int v \ dx - \int \left( \frac{du}{dx} \int v \ dx ight) dx$ ધારો કે $u = x$ અને $v = \frac{1}{2} \sec^{2} \frac{x}{2}$ $\Rightarrow I_{1} = x 	an \frac{x}{2} - \int 1 \cdot 	an \frac{x}{2} dx$ $\Rightarrow I_{1} = x 	an \frac{x}{2} - 2 \log_{e} \left| \sec \frac{x}{2} ight| + c_{1}$ $\Rightarrow I_{1} = x 	an \frac{x}{2} - \log_{e} \left| \sec^{2} \frac{x}{2} ight| + c_{1}$ હવે,$I_{2} = \int \frac{\sin x}{1 + \cos x} dx$ ધારો કે $1 + \cos x = t \Rightarrow - \sin x \ dx = dt$ $\Rightarrow I_{2} = - \int \frac{1}{t} dt = - \log_{e} |t| + c_{2} = - \log_{e} |1 + \cos x| + c_{2}$ $\Rightarrow I_{2} = - \log_{e} \left| 2 \cos^{2} \frac{x}{2} ight| + c_{2} = - \log_{e} 2 - \log_{e} \left| \cos^{2} \frac{x}{2} ight| + c_{2} = \log_{e} \left| \sec^{2} \frac{x}{2} ight| + c_{3}$ $I_{1}$ અને $I_{2}$ નો સરવાળો કરતા: $I = x 	an \frac{x}{2} - \log_{e} \left| \sec^{2} \frac{x}{2} ight| + c_{1} + \log_{e} \left| \sec^{2} \frac{x}{2} ight| + c_{3}$ $\Rightarrow I = x 	an \frac{x}{2} + c$

$\int \frac{x + \sin x}{1 + \cos x} dx =$

Similar Questions

$\int \frac{1+\sqrt{3} \cot x}{1-\sqrt{3} \cot x} d x=$

$\int \frac{dx}{(1+\sqrt{x}) \sqrt{x-x^2}} = $

$\int {\frac{{2x + 5}}{{\sqrt {7 - 6x - {x^2}} }}dx} = A\sqrt {7 - 6x - {x^2}} + B\,{\sin ^{ - 1}}\left( {\frac{{x + 3}}{4}} \right) + C$ (જ્યાં $C$ એ સંકલનનો અચળાંક છે),તો ક્રમયુક્ત જોડ $(A, B)$ બરાબર છે

$\int \frac{x^2}{(x\sin x + \cos x)^2} \, dx = $

$\frac{e^{-\pi/4} + \int_0^{\pi/4} e^{-x} \tan^{50} x \, dx}{\int_0^{\pi/4} e^{-x} (\tan^{49} x + \tan^{51} x) \, dx}$ ની કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module