Evaluate the definite integral int_{0}^{frac{ | vedclass

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

Question

(B) Let $I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x$. Divide the numerator and denominator by $\cos^{4} x$: $I = \

Vedclass · Accepted Answer

$\frac{\pi}{8}$

Vedclass · Answer

(B) Let $I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x$. Divide the numerator and denominator by $\cos^{4} x$: $I = \int_{0}^{\frac{\pi}{4}} \frac{	an x \sec^{2} x}{1+	an^{4} x} d x$. Let $	an^{2} x = t$. Then $2 	an x \sec^{2} x d x = d t$,which implies $	an x \sec^{2} x d x = \frac{1}{2} d t$. When $x = 0$,$t = 	an^{2} 0 = 0$. When $x = \frac{\pi}{4}$,$t = 	an^{2} \frac{\pi}{4} = 1$. Substituting these into the integral: $I = \frac{1}{2} \int_{0}^{1} \frac{d t}{1+t^{2}}$. Using the standard integral $\int \frac{1}{1+t^{2}} d t = 	an^{-1} t$: $I = \frac{1}{2} [	an^{-1} t]_{0}^{1} = \frac{1}{2} (	an^{-1} 1 - 	an^{-1} 0) = \frac{1}{2} (\frac{\pi}{4} - 0) = \frac{\pi}{8}$.

Evaluate the definite integral $\int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x$.

Similar Questions

The greatest value of the function $F(x) = \int_1^x {|t| \, dt}$ on the interval $\left[ -\frac{1}{2}, \frac{1}{2} \right]$ is:

Prove that $\int_{0}^{\frac{\pi}{2}} \sin^{3} x \, dx = \frac{2}{3}$.

$\int_0^{\pi / 4} \sqrt{1-\sin 2 x} \,d x =$

$\int_0^\infty {\frac{{x\,dx}}{{(1 + x)(1 + {x^2})}}} = $

Evaluate the definite integral: $\int_0^{\pi /8} \frac{\sec^2 2x}{2} \, dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module