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

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

Question

(A) Let $I = \int_0^{\pi / 4} (\sqrt{	an x} + \sqrt{\cot x}) \, dx$. $I = \int_0^{\pi / 4} \left( \frac{\sqrt{\sin x}}{\sqrt{\cos x}} + \frac{\sqrt{\

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int_0^{\pi / 4} (\sqrt{	an x} + \sqrt{\cot x}) \, dx$. $I = \int_0^{\pi / 4} \left( \frac{\sqrt{\sin x}}{\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x}} ight) \, dx$. $I = \int_0^{\pi / 4} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \, dx$. Multiply numerator and denominator by $\sqrt{2}$: $I = \int_0^{\pi / 4} \frac{\sqrt{2}(\sin x + \cos x)}{\sqrt{2 \sin x \cos x}} \, dx = \sqrt{2} \int_0^{\pi / 4} \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} \, dx$. Let $t = \sin x - \cos x$, then $dt = (\cos x + \sin x) \, dx$. When $x = 0$, $t = -1$. When $x = \pi/4$, $t = 0$. $I = \sqrt{2} \int_{-1}^0 \frac{dt}{\sqrt{1 - t^2}} = \sqrt{2} [\sin^{-1} t]_{-1}^0$. $I = \sqrt{2} [\sin^{-1}(0) - \sin^{-1}(-1)] = \sqrt{2} [0 - (-\pi/2)] = \sqrt{2} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt{2}}$.

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

Similar Questions

$\int_{-2}^2 |[x]| \, dx$ is equal to

If $\int_{0}^{\sqrt{3}} \frac{15 x^{3}}{\sqrt{1+x^{2}+\sqrt{(1+x^{2})^{3}}}} dx = \alpha \sqrt{2} + \beta \sqrt{3}$,where $\alpha, \beta$ are integers,then $\alpha + \beta$ is equal to.

The integral $\int_{0}^{1} \frac{1}{7^{\left[\frac{1}{x}\right]}} dx$,where $[.]$ denotes the greatest integer function,is equal to:

Evaluate the following definite integral as a limit of sums:
$\int_{0}^{5}(x+1) d x$

If $I = \int_0^{100\pi} \sqrt{1 - \cos 2x} \, dx$,then the value of $I$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module