intlimits_{frac{pi }{6}}^{frac{5pi }{6}} {lef | vedclass

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

Question

(B) Let $I = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \frac{1}{2} [9\sin^2 	heta - (1 + 2\sin 	heta + \sin^2 	heta)] \, d	heta$ $I = \frac{1}{2} \int

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(B) Let $I = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \frac{1}{2} [9\sin^2 	heta - (1 + 2\sin 	heta + \sin^2 	heta)] \, d	heta$ $I = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} (8\sin^2 	heta - 2\sin 	heta - 1) \, d	heta$ Using $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$,we get: $I = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} [8(\frac{1 - \cos 2	heta}{2}) - 2\sin 	heta - 1] \, d	heta$ $I = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} (4 - 4\cos 2	heta - 2\sin 	heta - 1) \, d	heta$ $I = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} (3 - 4\cos 2	heta - 2\sin 	heta) \, d	heta$ $I = \frac{1}{2} [3	heta - 2\sin 2	heta + 2\cos 	heta]_{\frac{\pi}{6}}^{\frac{5\pi}{6}}$ $I = \frac{1}{2} [ (3(\frac{5\pi}{6}) - 2\sin(\frac{5\pi}{3}) + 2\cos(\frac{5\pi}{6})) - (3(\frac{\pi}{6}) - 2\sin(\frac{\pi}{3}) + 2\cos(\frac{\pi}{6})) ]$ $I = \frac{1}{2} [ (\frac{5\pi}{2} - 2(-\frac{\sqrt{3}}{2}) + 2(-\frac{\sqrt{3}}{2})) - (\frac{\pi}{2} - 2(\frac{\sqrt{3}}{2}) + 2(\frac{\sqrt{3}}{2})) ]$ $I = \frac{1}{2} [ (\frac{5\pi}{2} + \sqrt{3} - \sqrt{3}) - (\frac{\pi}{2} - \sqrt{3} + \sqrt{3}) ]$ $I = \frac{1}{2} [ \frac{5\pi}{2} - \frac{\pi}{2} ] = \frac{1}{2} [ 2\pi ] = \pi$.

$\int\limits_{\frac{\pi }{6}}^{\frac{5\pi }{6}} {\left( {\frac{1}{2}{{(3\sin \theta )}^2} - \frac{1}{2}{{(1 + \sin \theta )}^2}} \right)\,d\theta } $

Similar Questions

The value of $\int_0^2 [x^2] dx$ is (where $[x]$ denotes the greatest integer function not greater than $x$)

If $I_n = \int_0^{\pi/4} \tan^n \theta \, d\theta$ for $n = 1, 2, 3, \ldots$,then $I_{n-1} + I_{n+1}$ is equal to

$\int_0^\pi \frac{\cos x}{\sqrt{1-\sin ^2 x}} d x=$

If $\int_{1}^{2} \frac{dx}{(x^2 - 2x + 4)^{3/2}} = \frac{k}{k+5}$,then $k$ is equal to

The limit of the area under the curve $y = e^{-x}$ from $x = 0$ to $x = h$ as $h \rightarrow \infty$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module