If int {frac{{{e^x}(1 + sin x)}}{{1 + cos x}} | 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 the trigonometric identities $1 + \cos x = 2\cos

Vedclass · Accepted Answer

$	an \frac{x}{2}$

Vedclass · Answer

(C) We are given the integral $I = \int {e^x \left( \frac{1 + \sin x}{1 + \cos x} ight)} dx$. Using the trigonometric identities $1 + \cos x = 2\cos^2(x/2)$ and $\sin x = 2\sin(x/2)\cos(x/2)$,we get: $I = \int e^x \left[ \frac{1 + 2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} ight] dx$ $I = \int e^x \left[ \frac{1}{2\cos^2(x/2)} + \frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} ight] dx$ $I = \int e^x \left[ \frac{1}{2}\sec^2(x/2) + 	an(x/2) ight] dx$. Let $f(x) = 	an(x/2)$. Then $f'(x) = \frac{1}{2}\sec^2(x/2)$. Since the integral is of the form $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$,we have: $I = e^x 	an(x/2) + c$. Comparing this with the given expression ${e^x}f(x) + c$,we find $f(x) = 	an(x/2)$.

If $\int {\frac{{{e^x}(1 + \sin x)}}{{1 + \cos x}}} dx = {e^x}f(x) + c$,then $f(x) = $

Similar Questions

If $\int {\frac{{{x^2} - x + 1}}{{{x^2} + 1}}{e^{{{\cot }^{ - 1}}x}}dx = A(x) {e^{{{\cot }^{ - 1}}x}} + C}$,then $A(x)$ is equal to

$\int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx = $ . . . . . . $+ c$.

$\int {{e^x}\left[ {{{\sin }^{ - 1}}\frac{x}{a} + \frac{1}{{\sqrt {{a^2} - {x^2}} }}} \right]dx = }$

$\int \log x \cdot [\log (ex)]^{-2} dx = . . . . . .$

$\int {{e^x}(1 + \tan x + {{\tan }^2}x)\,dx = } $

Vedclass Test Series

Exam Paper Generator

Online Exam Module