If int e^x left( frac{1 - sin x}{1 - cos x} r | vedclass

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

Question

(C) We are given the integral $I = \int e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx$. Using trigonometric identities $\sin x = 2 \sin \frac{x}

Vedclass · Accepted Answer

$-e^x \cot \left( \frac{x}{2} \right)$

Vedclass · Answer

(C) We are given the integral $I = \int e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx$. Using trigonometric identities $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1 - \cos x = 2 \sin^2 \frac{x}{2}$,we get: $I = \int e^x \left( \frac{1 - 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \right) dx$ $I = \int e^x \left( \frac{1}{2 \sin^2 \frac{x}{2}} - \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \right) dx$ $I = \int e^x \left( \frac{1}{2} \operatorname{cosec}^2 \frac{x}{2} - \cot \frac{x}{2} \right) dx$ Let $g(x) = -\cot \frac{x}{2}$. Then $g'(x) = -(-\operatorname{cosec}^2 \frac{x}{2} \cdot \frac{1}{2}) = \frac{1}{2} \operatorname{cosec}^2 \frac{x}{2}$. Since the integral is of the form $\int e^x (g(x) + g'(x)) dx = e^x g(x) + c$,we have: $I = e^x \left( -\cot \frac{x}{2} \right) + c = -e^x \cot \frac{x}{2} + c$. Thus,$f(x) = -e^x \cot \left( \frac{x}{2} \right)$.

If $\int e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx = f(x) + \text{constant}$,then $f(x)$ is equal to

Similar Questions

$\int 2^{x} [f^{\prime}(x) + f(x) \log 2] \, dx$ is equal to

$\int {{e^{\sin x}}\left( {\sin x + {{\sec }^2}x} \right)} \,dx$ is equal to

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

$\int e^x \tan x(1+\tan x) \, dx = $ . . . . . . $+ C$.

Let $f(x) = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x}$. Then find $\int e^x(f(x) + f'(x)) dx$ (where $c$ is the constant of integration).

Vedclass Test Series

Exam Paper Generator

Online Exam Module