int_0^{pi / 2} frac{sin x}{1+cos x+sin x} d x | vedclass

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

Question

(A) Let $I = \int_0^{\pi / 2} \frac{\sin x}{1+\cos x+\sin x} d x$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get: $I = \int_0^{\pi

Vedclass · Accepted Answer

$\frac{\pi}{4}-\frac{1}{2} \log 2$

Vedclass · Answer

(A) Let $I = \int_0^{\pi / 2} \frac{\sin x}{1+\cos x+\sin x} d x$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get: $I = \int_0^{\pi / 2} \frac{\sin(\pi/2 - x)}{1+\cos(\pi/2 - x)+\sin(\pi/2 - x)} dx = \int_0^{\pi / 2} \frac{\cos x}{1+\sin x+\cos x} dx$. Adding the two expressions for $I$: $2I = \int_0^{\pi / 2} \frac{\sin x + \cos x}{1+\sin x+\cos x} dx$. $2I = \int_0^{\pi / 2} \frac{(1+\sin x+\cos x) - 1}{1+\sin x+\cos x} dx = \int_0^{\pi / 2} (1 - \frac{1}{1+\sin x+\cos x}) dx$. $2I = [x]_0^{\pi / 2} - \int_0^{\pi / 2} \frac{1}{1+2\sin(x/2)\cos(x/2)+2\cos^2(x/2)-1} dx$. $2I = \frac{\pi}{2} - \int_0^{\pi / 2} \frac{1}{2\cos^2(x/2)(	an(x/2)+1)} dx = \frac{\pi}{2} - \frac{1}{2} \int_0^{\pi / 2} \frac{\sec^2(x/2)}{	an(x/2)+1} dx$. Let $u = 	an(x/2)$,then $du = \frac{1}{2} \sec^2(x/2) dx$. $2I = \frac{\pi}{2} - \int_0^1 \frac{1}{u+1} du = \frac{\pi}{2} - [\log|u+1|]_0^1 = \frac{\pi}{2} - \log 2$. Therefore,$I = \frac{\pi}{4} - \frac{1}{2} \log 2$.

$\int_0^{\pi / 2} \frac{\sin x}{1+\cos x+\sin x} d x=$

Similar Questions

The value of $\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x$ is

$\int_0^{\frac{\pi}{2}} \frac{300 \sin x+100 \cos x}{\sin x+\cos x} \,dx = \ldots$ (in $\pi$)

If $I$ is the greatest of the definite integrals
${I_1} = \int_0^1 {{e^{ - x}}{{\cos }^2}x\,dx} , \,\, {I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x\,dx$
${I_3} = \int_0^1 {{e^{ - {x^2}}}dx} ,\,\,{I_4} = \int_0^1 {{e^{ - {x^2}/2}}dx} ,$ then

The value of $\int_a^{a + (\pi /2)} (\sin^4 x + \cos^4 x) \, dx$ is

$\int_{0}^{\frac{\pi}{2}} \log \left[\sqrt{\frac{1-\cos 2x}{1+\cos 2x}}\right] dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module