int frac{10^{frac{x}{2}}}{sqrt{10^{-x}-10^x}} | vedclass

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

Question

(B) Let $I = \int \frac{10^{\frac{x}{2}}}{\sqrt{10^{-x}-10^x}} dx$ We can rewrite the denominator as: $\sqrt{10^{-x}-10^x} = \sqrt{\frac{1}{10^x} - 10

Vedclass · Accepted Answer

$\frac{1}{\log 10} \sin ^{-1}\left(10^x\right)+c$

Vedclass · Answer

(B) Let $I = \int \frac{10^{\frac{x}{2}}}{\sqrt{10^{-x}-10^x}} dx$ We can rewrite the denominator as: $\sqrt{10^{-x}-10^x} = \sqrt{\frac{1}{10^x} - 10^x} = \sqrt{\frac{1-(10^x)^2}{10^x}} = \frac{\sqrt{1-(10^x)^2}}{10^{\frac{x}{2}}}$ Substituting this into the integral: $I = \int \frac{10^{\frac{x}{2}}}{\frac{\sqrt{1-(10^x)^2}}{10^{\frac{x}{2}}}} dx = \int \frac{10^x}{\sqrt{1-(10^x)^2}} dx$ Let $t = 10^x$. Then $dt = 10^x (\log 10) dx$,which implies $10^x dx = \frac{dt}{\log 10}$. Substituting $t$ into the integral: $I = \frac{1}{\log 10} \int \frac{dt}{\sqrt{1-t^2}} = \frac{1}{\log 10} \sin^{-1}(t) + c$ Replacing $t$ with $10^x$: $I = \frac{1}{\log 10} \sin^{-1}(10^x) + c$

$\int \frac{10^{\frac{x}{2}}}{\sqrt{10^{-x}-10^x}} dx=$

Similar Questions

$\int \frac{d x}{(x+100) \sqrt{x+99}}=f(x)+c \Rightarrow f(x)$

Evaluate $\int \sin (\sqrt{k}) \, dk$ on $(0, \infty)$.

$\int \sin^5 x \cos^4 x \, dx = $

$\int \sqrt{\operatorname{cosec} x - \sin x} \, dx$ is equal to

$\int \frac{x^2}{(9 - x^2)^{3/2}} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module