The value of int_a^{a + (pi /2)} (sin^4 x + c | vedclass

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

Question

(C) Let $I = \int_a^{a + (\pi /2)} (\sin^4 x + \cos^4 x) \, dx$. Since the integrand $f(x) = \sin^4 x + \cos^4 x$ is a periodic function with period $

Vedclass · Accepted Answer

$\frac{3\pi}{8}$

Vedclass · Answer

(C) Let $I = \int_a^{a + (\pi /2)} (\sin^4 x + \cos^4 x) \, dx$. Since the integrand $f(x) = \sin^4 x + \cos^4 x$ is a periodic function with period $\frac{\pi}{2}$,the integral over any interval of length $\frac{\pi}{2}$ is constant and equal to the integral over $[0, \pi/2]$. Thus,$I = \int_0^{\pi/2} (\sin^4 x + \cos^4 x) \, dx$. Using the identity $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}(2\sin x \cos x)^2 = 1 - \frac{1}{2}\sin^2(2x)$. Therefore,$I = \int_0^{\pi/2} (1 - \frac{1}{2}\sin^2(2x)) \, dx = \int_0^{\pi/2} 1 \, dx - \frac{1}{2} \int_0^{\pi/2} \sin^2(2x) \, dx$. $I = [x]_0^{\pi/2} - \frac{1}{2} \int_0^{\pi/2} \frac{1 - \cos(4x)}{2} \, dx = \frac{\pi}{2} - \frac{1}{4} [x - \frac{\sin(4x)}{4}]_0^{\pi/2}$. $I = \frac{\pi}{2} - \frac{1}{4} (\frac{\pi}{2} - 0) = \frac{\pi}{2} - \frac{\pi}{8} = \frac{3\pi}{8}$.

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

Similar Questions

$\int\limits_{\frac{1}{2}}^{3\frac{1}{2}} {\left\{ {\frac{1}{2}\,\left( {|x - 3| + |1 - x| - 4} \right)} \right\}\,dx} $ equals: Where $\{*\}$ denotes the fractional part function.

The value of $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^{-x}} \,dx$ is equal to

If $n$ is a positive integer and $[x]$ is the greatest integer not exceeding $x$,then $\int_0^n {\{x - [x]\} \,dx}$ equals

If $I = \int_0^\pi x \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$,then $[I] = \ldots$. Here,$[.]$ denotes the greatest integer function.

If $f(a+b+1-x)=f(x)$ for all $x,$ where $a$ and $b$ are fixed positive real numbers,then $\frac{1}{a+b} \int_{a}^{b} x(f(x)+f(x+1)) dx$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module