int frac{x+sin x}{1+cos x} d x is equal to | vedclass

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

Question

(D) Let $I = \int \frac{x+\sin x}{1+\cos x} dx$. Using the identities $\sin x = 2 \sin(x/2) \cos(x/2)$ and $1+\cos x = 2 \cos^2(x/2)$,we get: $I = \in

Vedclass · Accepted Answer

$x 	an \left(\frac{x}{2}ight)+c$

Vedclass · Answer

(D) Let $I = \int \frac{x+\sin x}{1+\cos x} dx$. Using the identities $\sin x = 2 \sin(x/2) \cos(x/2)$ and $1+\cos x = 2 \cos^2(x/2)$,we get: $I = \int \frac{x + 2 \sin(x/2) \cos(x/2)}{2 \cos^2(x/2)} dx$ $I = \int \left( \frac{x}{2 \cos^2(x/2)} + \frac{2 \sin(x/2) \cos(x/2)}{2 \cos^2(x/2)} ight) dx$ $I = \int \left( \frac{1}{2} x \sec^2(x/2) + 	an(x/2) ight) dx$ Now,apply integration by parts to the first term $\int \frac{1}{2} x \sec^2(x/2) dx$: Let $u = x$ and $dv = \frac{1}{2} \sec^2(x/2) dx$. Then $du = dx$ and $v = 	an(x/2)$. Using $\int u dv = uv - \int v du$: $\int \frac{1}{2} x \sec^2(x/2) dx = x 	an(x/2) - \int 	an(x/2) dx$ Substituting this back into the expression for $I$: $I = x 	an(x/2) - \int 	an(x/2) dx + \int 	an(x/2) dx + c$ $I = x 	an(x/2) + c$

$\int \frac{x+\sin x}{1+\cos x} d x$ is equal to

Similar Questions

Let for $f(x)=7 \tan^8 x + 7 \tan^6 x - 3 \tan^4 x - 3 \tan^2 x$,$I_1 = \int_0^{\pi/4} f(x) \, dx$ and $I_2 = \int_0^{\pi/4} x f(x) \, dx$. Then $7 I_1 + 12 I_2$ is equal to:

If $\int {\frac{1}{{x + {x^5}}}dx = f(x) + c} $,then the value of $\int {\frac{{{x^4}}}{{x + {x^5}}}dx} $ is

$\int \frac{d x}{(x-2) \sqrt{x^2-3 x+5}} =$

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module