int(sqrt{tan x}+sqrt{cot x}) d x= | vedclass

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

Question

(D) Let $I = \int(\sqrt{	an x}+\sqrt{\cot x}) d x$. Expressing in terms of $\sin x$ and $\cos x$: $I = \int \left(\sqrt{\frac{\sin x}{\cos x}} + \sqr

Vedclass · Accepted Answer

$\sqrt{2} \sin ^{-1}(\sin x-\cos x)+c$

Vedclass · Answer

(D) Let $I = \int(\sqrt{	an x}+\sqrt{\cot x}) d x$. Expressing in terms of $\sin x$ and $\cos x$: $I = \int \left(\sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}}ight) d x = \int \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} d x$. Multiply and divide by $\sqrt{2}$: $I = \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{2 \sin x \cos x}} d x$. Since $1 - (\sin x - \cos x)^2 = 1 - (\sin^2 x + \cos^2 x - 2 \sin x \cos x) = 2 \sin x \cos x$: $I = \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} d x$. Let $t = \sin x - \cos x$, then $dt = (\cos x + \sin x) d x$. Substituting these into the integral: $I = \sqrt{2} \int \frac{dt}{\sqrt{1 - t^2}} = \sqrt{2} \sin^{-1}(t) + c$. Substituting back $t = \sin x - \cos x$: $I = \sqrt{2} \sin^{-1}(\sin x - \cos x) + c$.

$\int(\sqrt{\tan x}+\sqrt{\cot x}) d x=$

Similar Questions

If $f(x) = \int \csc^5 x \ dx$,then $f(\frac{\pi}{4}) = $

$\int \frac{dx}{x^2 \sqrt{4+x^2}}$ is equal to

For $n \geq 2$,if $I_n = \int \sec^n x \, dx$,then $I_4 - \frac{2}{3} I_2 =$

If $\int \frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x} d x=A \log |5 \cos x+4 \sin x|+B x+c$,then $A$ and $B$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module