Evaluate the integral: int_0^pi left(cos^2 le | vedclass

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

Question

(C) Let $I = \int_0^\pi \left(\cos^2 \left(\frac{3\pi}{8} - \frac{x}{4}\right) - \cos^2 \left(\frac{11\pi}{8} + \frac{x}{4}\right)\right) dx$. Using t

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(C) Let $I = \int_0^\pi \left(\cos^2 \left(\frac{3\pi}{8} - \frac{x}{4}\right) - \cos^2 \left(\frac{11\pi}{8} + \frac{x}{4}\right)\right) dx$. Using the identity $\cos^2 A - \cos^2 B = \sin(B-A) \sin(B+A)$,we have: $A = \frac{3\pi}{8} - \frac{x}{4}$ and $B = \frac{11\pi}{8} + \frac{x}{4}$. $B - A = \left(\frac{11\pi}{8} + \frac{x}{4}\right) - \left(\frac{3\pi}{8} - \frac{x}{4}\right) = \frac{8\pi}{8} + \frac{2x}{4} = \pi + \frac{x}{2}$. $B + A = \left(\frac{11\pi}{8} + \frac{x}{4}\right) + \left(\frac{3\pi}{8} - \frac{x}{4}\right) = \frac{14\pi}{8} = \frac{7\pi}{4}$. Thus,the integrand becomes $\sin(\pi + \frac{x}{2}) \sin(\frac{7\pi}{4}) = (-\sin \frac{x}{2}) \cdot (-\frac{1}{\sqrt{2}}) = \frac{1}{\sqrt{2}} \sin \frac{x}{2}$. $I = \int_0^\pi \frac{1}{\sqrt{2}} \sin \frac{x}{2} dx = \frac{1}{\sqrt{2}} \left[ -2 \cos \frac{x}{2} \right]_0^\pi$. $I = \frac{-2}{\sqrt{2}} (\cos \frac{\pi}{2} - \cos 0) = -\sqrt{2} (0 - 1) = \sqrt{2}$.

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

Evaluate the integral: $\int_0^\pi \left(\cos^2 \left(\frac{3\pi}{8} - \frac{x}{4}\right) - \cos^2 \left(\frac{11\pi}{8} + \frac{x}{4}\right)\right) dx$

Similar Questions

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

If $\int_{\log _{e} 2}^{x} (e^{t}-1)^{-1} dt = \log _{e} \frac{3}{2}$, then the value of $x$ is

$\int_{-2}^{2} |1 - x^2| \, dx = $

The integral $\int_0^{\frac{\pi}{4}} \frac{136 \sin x}{3 \sin x+5 \cos x} dx$ is equal to :

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module