int e^x left( frac{1 + sin x}{1 + cos x} righ | vedclass

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

Question

(B) We know that $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$. Given integral is $I = \int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx$. Using

Vedclass · Accepted Answer

$e^x 	an \frac{x}{2}$

Vedclass · Answer

(B) We know that $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$. Given integral is $I = \int e^x \left( \frac{1 + \sin x}{1 + \cos x} ight) dx$. Using trigonometric identities: $1 + \sin x = 1 + 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1 + \cos x = 2 \cos^2 \frac{x}{2}$. So,$\frac{1 + \sin x}{1 + \cos x} = \frac{1 + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} = \frac{1}{2} \sec^2 \frac{x}{2} + 	an \frac{x}{2}$. Let $f(x) = 	an \frac{x}{2}$. Then $f'(x) = \sec^2 \frac{x}{2} \cdot \frac{1}{2} = \frac{1}{2} \sec^2 \frac{x}{2}$. Thus,the integral becomes $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C = e^x 	an \frac{x}{2} + C$. Therefore,the correct option is $B$.

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

Similar Questions

The midpoint of the chord $4x - 3y = 5$ of the hyperbola $2x^2 - 3y^2 = 12$ is

If the circle $x^2+y^2+6x-2y+k=0$ bisects the circumference of the circle $x^2+y^2+2x-6y-15=0$,then $k=$

If $S$ and $S^{\prime}$ are the foci of an ellipse,$B$ is one end of the minor axis and $\angle S B S^{\prime} = 90^{\circ}$,then the eccentricity of that ellipse is

The work done in moving an object from the origin to a point whose position vector is $r = 3 \hat{i} + 2 \hat{j} - 5 \hat{k}$ by a force $F = 2 \hat{i} - \hat{j} - \hat{k}$ is:

If the equation $\cos^{4} \theta + \sin^{4} \theta + \lambda = 0$ has real solutions for $\theta$,then $\lambda$ lies in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module