int_0^{pi /4} [sqrt{tan x} + sqrt{cot x}] , d | vedclass

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

Question

(C) Let $I = \int_0^{\pi /4} [\sqrt{	an x} + \sqrt{\cot x}] \, dx$. We can rewrite the integrand as: $I = \int_0^{\pi /4} \left( \sqrt{\frac{\sin x}{

Vedclass · Accepted Answer

$\frac{\pi}{\sqrt{2}}$

Vedclass · Answer

(C) Let $I = \int_0^{\pi /4} [\sqrt{	an x} + \sqrt{\cot x}] \, dx$. We can rewrite the integrand as: $I = \int_0^{\pi /4} \left( \sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}} ight) dx = \int_0^{\pi /4} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \, dx$. Multiply the numerator and denominator by $\sqrt{2}$: $I = \sqrt{2} \int_0^{\pi /4} \frac{\sin x + \cos x}{\sqrt{2 \sin x \cos x}} \, dx = \sqrt{2} \int_0^{\pi /4} \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} \, dx$. Let $t = \sin x - \cos x$. Then $dt = (\cos x + \sin x) \, dx$. When $x = 0$,$t = \sin 0 - \cos 0 = -1$. When $x = \pi / 4$,$t = \sin(\pi / 4) - \cos(\pi / 4) = 0$. Substituting these into the integral: $I = \sqrt{2} \int_{-1}^0 \frac{dt}{\sqrt{1 - t^2}} = \sqrt{2} [\sin^{-1} t]_{-1}^0$. $I = \sqrt{2} [\sin^{-1}(0) - \sin^{-1}(-1)] = \sqrt{2} [0 - (-\pi / 2)] = \sqrt{2} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt{2}}$.

$\int_0^{\pi /4} [\sqrt{\tan x} + \sqrt{\cot x}] \, dx$ equals

Similar Questions

If $[x]$ is the greatest integer function,then $\int_0^5 [x] \, dx =$

If $\int_0^b \frac{dx}{1+x^2} = \int_b^{\infty} \frac{dx}{1+x^2}$,then $b$ is equal to

$A$ minimum value of $\int_0^x t e^{t^2} d t$ is

If the real part of the complex number $(1-\cos \theta+2 i \sin \theta)^{-1}$ is $\frac{1}{5}$ for $\theta \in(0, \pi)$,then the value of the integral $\int_{0}^{\theta} \sin x \,dx$ is equal to:

$\int_1^e \frac{1}{x} \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module