int_0^{pi /4} frac{dx}{cos^4 x - cos^2 x sin^ | vedclass

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

Question

(A) Let $I = \int_0^{\pi /4} \frac{dx}{\cos^4 x - \cos^2 x \sin^2 x + \sin^4 x}$.Divide the numerator and denominator by $\cos^4 x$:$I = \int_0^{\pi /

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(A) Let $I = \int_0^{\pi /4} \frac{dx}{\cos^4 x - \cos^2 x \sin^2 x + \sin^4 x}$.Divide the numerator and denominator by $\cos^4 x$:$I = \int_0^{\pi /4} \frac{\sec^4 x dx}{1 - 	an^2 x + 	an^4 x} = \int_0^{\pi /4} \frac{(1 + 	an^2 x) \sec^2 x dx}{1 - 	an^2 x + 	an^4 x}$.Substitute $	an x = t$,so $\sec^2 x dx = dt$. When $x = 0, t = 0$ and when $x = \pi/4, t = 1$.$I = \int_0^1 \frac{1 + t^2}{t^4 - t^2 + 1} dt = \int_0^1 \frac{\frac{1}{t^2} + 1}{t^2 - 1 + \frac{1}{t^2}} dt = \int_0^1 \frac{1 + \frac{1}{t^2}}{(t - \frac{1}{t})^2 + 1} dt$.Let $u = t - \frac{1}{t}$,then $du = (1 + \frac{1}{t^2}) dt$. As $t 	o 0^+, u 	o -\infty$ and as $t 	o 1, u 	o 0$.$I = \int_{-\infty}^0 \frac{du}{u^2 + 1} = [	an^{-1} u]_{-\infty}^0 = 	an^{-1}(0) - 	an^{-1}(-\infty) = 0 - (-\frac{\pi}{2}) = \frac{\pi}{2}$.

$\int_0^{\pi /4} \frac{dx}{\cos^4 x - \cos^2 x \sin^2 x + \sin^4 x} = $

Similar Questions

If the value of the integral $\int_{0}^{5} \frac{x+[x]}{e^{x-[x]}} \,dx = \alpha e^{-1} + \beta$,where $\alpha, \beta \in R, 5\alpha + 6\beta = 0$,and $[x]$ denotes the greatest integer less than or equal to $x$; then the value of $(\alpha + \beta)^{2}$ is equal to:

$\int_{-1}^{2} {|x|\,dx} =$ (in $/2$)

$\int_{ - \pi /4}^{\pi /2} {{e^{ - x}}\sin x\,dx} = $

Evaluate the definite integral $\int_{2}^{3} \frac{d x}{x^{2}-1}$.

Let $\alpha \in (0,1)$ and $\beta = \log_{e}(1-\alpha)$. Let $P_n(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots + \frac{x^n}{n}$ for $x \in (0,1)$. Then the integral $\int_{0}^{\alpha} \frac{t^{50}}{1-t} dt$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module