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

(A) Let $I = \int (\sqrt{	an x} + \sqrt{\cot x}) dx = \int \frac{	an x + 1}{\sqrt{	an x}} dx$. Substitute $	an x = t^2$,then $\sec^2 x dx = 2t dt$

Vedclass · Accepted Answer

$\sqrt{2} 	an^{-1}\left(\frac{	an x-1}{\sqrt{2 	an x}}ight)+c$

Vedclass · Answer

(A) Let $I = \int (\sqrt{	an x} + \sqrt{\cot x}) dx = \int \frac{	an x + 1}{\sqrt{	an x}} dx$. Substitute $	an x = t^2$,then $\sec^2 x dx = 2t dt$. Since $\sec^2 x = 1 + 	an^2 x = 1 + t^4$,we have $dx = \frac{2t}{1+t^4} dt$. Thus,$I = \int \frac{t^2 + 1}{t} \cdot \frac{2t}{1+t^4} dt = 2 \int \frac{t^2 + 1}{t^4 + 1} dt$. Divide numerator and denominator by $t^2$: $I = 2 \int \frac{1 + 1/t^2}{t^2 + 1/t^2} dt = 2 \int \frac{1 + 1/t^2}{(t - 1/t)^2 + 2} dt$. Let $u = t - 1/t$,then $du = (1 + 1/t^2) dt$. $I = 2 \int \frac{du}{u^2 + (\sqrt{2})^2} = 2 \cdot \frac{1}{\sqrt{2}} 	an^{-1}\left(\frac{u}{\sqrt{2}}ight) + c = \sqrt{2} 	an^{-1}\left(\frac{t - 1/t}{\sqrt{2}}ight) + c$. Substituting $t = \sqrt{	an x}$,we get $I = \sqrt{2} 	an^{-1}\left(\frac{	an x - 1}{\sqrt{2 	an x}}ight) + c$.

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

Similar Questions

If $\int(1+x-x^{-1}) e^{x+x^{-1}} dx = f(x)+c$, then $f(1)-f(-1)=$

Assertion $(A)$: If $I_n = \int \cot^n x \, dx$,then $I_6 + I_4 = \frac{-\cot^5 x}{5}$.
Reason $(R)$: $\int \cot^n x \, dx = \frac{-\cot^{n-1} x}{n-1} - \int \cot^{n-2} x \, dx$.

If $\int \frac{x^2-x+2}{x^2+x+2} d x=x-\log (f(x))+\frac{2}{\sqrt{7}} \operatorname{Tan}^{-1}(g(x))+c$,then $f(-1)+\sqrt{7} g(-1)=$

Integrate the function: $\frac{x+2}{\sqrt{x^{2}-1}}$

Evaluate the integral: $\int \sqrt{4x(4x + 4)} \, dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module