int_{-pi}^{pi} frac{sin^4 x}{sin^4 x + cos^4 | vedclass

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

Question

(D) Let $I = \int_{-\pi}^{\pi} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx$. Since the integrand $f(x) = \frac{\sin^4 x}{\sin^4 x + \cos^4 x}$ is an ev

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(D) Let $I = \int_{-\pi}^{\pi} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx$. Since the integrand $f(x) = \frac{\sin^4 x}{\sin^4 x + \cos^4 x}$ is an even function,we have $I = 2 \int_{0}^{\pi} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx$. Using the property $\int_{0}^{2a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ if $f(2a-x) = f(x)$,we note that $f(\pi - x) = \frac{\sin^4(\pi - x)}{\sin^4(\pi - x) + \cos^4(\pi - x)} = \frac{\sin^4 x}{\sin^4 x + \cos^4 x} = f(x)$. Thus,$I = 2 	imes 2 \int_{0}^{\pi/2} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx = 4 \int_{0}^{\pi/2} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx$ ..... $(i)$. Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get $I = 4 \int_{0}^{\pi/2} \frac{\cos^4 x}{\cos^4 x + \sin^4 x} \, dx$ ..... $(ii)$. Adding $(i)$ and $(ii)$,we get $2I = 4 \int_{0}^{\pi/2} \frac{\sin^4 x + \cos^4 x}{\sin^4 x + \cos^4 x} \, dx = 4 \int_{0}^{\pi/2} 1 \, dx = 4 	imes \frac{\pi}{2} = 2\pi$. Therefore,$I = \pi$.

$\int_{-\pi}^{\pi} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx = $

Similar Questions

$\int_{2}^{4} \frac{\log(x^2)}{\log(x^2) + \log(36 - 12x + x^2)} \, dx = $

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

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

$\int_{-\pi}^{\pi} x^2 \sin x \, dx =$

Let $[t]$ denote the greatest integer $\leq t$. Then $\frac{2}{\pi} \int_{\pi/6}^{5\pi/6} (8[\operatorname{cosec} x] - 5[\cot x]) \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module