int_0^{pi /2} {left( {frac{theta }{{sin theta | vedclass

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

Question

(A) Let $I = \int_0^{\pi /2} {\left( {\frac{	heta }{{\sin 	heta }}} ight)^2 d	heta } $. Using integration by parts,let $u = 	heta^2$ and $dv = \

Vedclass · Accepted Answer

$\pi \log 2$

Vedclass · Answer

(A) Let $I = \int_0^{\pi /2} {\left( {\frac{	heta }{{\sin 	heta }}} ight)^2 d	heta } $. Using integration by parts,let $u = 	heta^2$ and $dv = \csc^2 	heta \, d	heta$. Then $du = 2	heta \, d	heta$ and $v = -\cot 	heta$. $I = [-	heta^2 \cot 	heta]_0^{\pi/2} + \int_0^{\pi/2} 2	heta \cot 	heta \, d	heta$. Evaluating the boundary term: $\lim_{	heta 	o \pi/2} (-	heta^2 \cot 	heta) = 0$ and $\lim_{	heta 	o 0} (-	heta^2 \cot 	heta) = \lim_{	heta 	o 0} (-	heta^2 \frac{\cos 	heta}{\sin 	heta}) = 0$. So,$I = 2 \int_0^{\pi/2} 	heta \cot 	heta \, d	heta$. Using integration by parts again,let $u = 	heta$ and $dv = \cot 	heta \, d	heta$. Then $du = d	heta$ and $v = \ln(\sin 	heta)$. $I = 2 [	heta \ln(\sin 	heta)]_0^{\pi/2} - 2 \int_0^{\pi/2} \ln(\sin 	heta) \, d	heta$. Since $\ln(\sin(\pi/2)) = \ln(1) = 0$ and $\lim_{	heta 	o 0} 	heta \ln(\sin 	heta) = 0$,the first term is $0$. We know that $\int_0^{\pi/2} \ln(\sin 	heta) \, d	heta = -\frac{\pi}{2} \ln 2$. Therefore,$I = 0 - 2(-\frac{\pi}{2} \ln 2) = \pi \ln 2$.

$\int_0^{\pi /2} {\left( {\frac{\theta }{{\sin \theta }}} \right)^2 d\theta = } $

Similar Questions

The integral $\int \frac{\left(1-\frac{1}{\sqrt{3}}\right)(\cos x-\sin x)}{\left(1+\frac{2}{\sqrt{3}} \sin 2 x\right)} d x$ is equal to

If $\int \frac{dx}{\sin^3 x + \cos^3 x} = A \log \left|\frac{\sqrt{2}+t}{\sqrt{2}-t}\right| + B \tan^{-1}(t) + c$,then $\left(\frac{B}{A}, t\right) =$

If $f(x) = \int \csc^5 x \ dx$,then $f(\frac{\pi}{4}) = $

If $\int \frac{2 e^x+3 e^{-x}}{3 e^x+4 e^{-x}} d x=A x+B \log \left(3 e^{2 x}+4\right)+C$,then values of $A$ and $B$ are respectively (where $C$ is a constant of integration.)

If $\int \frac{x-\sin x}{1+\cos x} dx = x \tan \left(\frac{x}{2}\right) + p \log \left|\sec \left(\frac{x}{2}\right)\right| + C$,then $p$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module