int_0^pi frac{1}{1+sin x} dx is equal to | vedclass

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

Question

(B) Let $I = \int_0^\pi \frac{1}{1+\sin x} dx$. Using the identity $\sin x = \frac{2 	an(x/2)}{1+	an^2(x/2)}$, we get: $I = \int_0^\pi \frac{1}{1+\f

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Let $I = \int_0^\pi \frac{1}{1+\sin x} dx$. Using the identity $\sin x = \frac{2 	an(x/2)}{1+	an^2(x/2)}$, we get: $I = \int_0^\pi \frac{1}{1+\frac{2 	an(x/2)}{1+	an^2(x/2)}} dx = \int_0^\pi \frac{1+	an^2(x/2)}{1+	an^2(x/2)+2 	an(x/2)} dx$. Since $1+	an^2(x/2) = \sec^2(x/2)$, we have: $I = \int_0^\pi \frac{\sec^2(x/2)}{(1+	an(x/2))^2} dx$. Let $t = 	an(x/2)$, then $dt = \frac{1}{2} \sec^2(x/2) dx$, which implies $\sec^2(x/2) dx = 2 dt$. As $x 	o 0$, $t 	o 0$. As $x 	o \pi$, $t 	o \infty$. Thus, $I = \int_0^\infty \frac{2 dt}{(1+t)^2} = 2 \left[ -\frac{1}{1+t} ight]_0^\infty = 2(0 - (-1)) = 2$.

$t \ (\text{in second})$	$0$	$2$	$4$	$6$	$8$	$10$
$v \ (\text{in m/s})$	$0$	$12$	$16$	$20$	$35$	$60$

$\int_0^\pi \frac{1}{1+\sin x} dx$ is equal to

Similar Questions

Let $f:(2, \infty) \rightarrow \mathbb{N}$ be defined by $f(x) =$ the largest prime factor of $[x]$. Then,$\int_{2}^{8} f(x) \, dx$ is equal to

Number of values of $x$ satisfying the equation $\int_{-1}^{x} (8t^2 + \frac{28}{3}t + 4) dt = \frac{(\frac{3}{2})x + 1}{\log_{(x+1)} \sqrt{x+1}}$.

$\int_{0.2}^{3.5} [x] \, dx = $ (where $[x]$ is the greatest integer function not greater than $x$)

The velocity of a particle which starts from rest is given by the following table. The total distance travelled (in metre) by the particle in $10 \ s$, using the Trapezoidal rule, is:
$t \ (\text{in second})$ $0$ $2$ $4$ $6$ $8$ $10$
$v \ (\text{in m/s})$ $0$ $12$ $16$ $20$ $35$ $60$

$\int_0^1 x \operatorname{Sin}^{-1} x \, dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module