int frac{a^x}{sqrt{1 - a^{2x}}} dx = | vedclass

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

Question

(A) Let $a^x = t$. Then,differentiating both sides with respect to $x$,we get $a^x \log_e a \, dx = dt$,which implies $a^x \, dx = \frac{dt}{\log_e a}

Vedclass · Accepted Answer

$\frac{1}{\log a} \sin^{-1}(a^x) + c$

Vedclass · Answer

(A) Let $a^x = t$. Then,differentiating both sides with respect to $x$,we get $a^x \log_e a \, dx = dt$,which implies $a^x \, dx = \frac{dt}{\log_e a}$.Substituting these into the integral:$\int \frac{a^x}{\sqrt{1 - a^{2x}}} dx = \int \frac{1}{\sqrt{1 - t^2}} \cdot \frac{dt}{\log_e a}$$= \frac{1}{\log_e a} \int \frac{dt}{\sqrt{1 - t^2}}$$= \frac{1}{\log_e a} \sin^{-1}(t) + c$$= \frac{\sin^{-1}(a^x)}{\log_e a} + c$.

$\int \frac{a^x}{\sqrt{1 - a^{2x}}} dx = $

Similar Questions

$\int \frac{d x}{e^{x}+e^{-x}}$ is equal to

$\int \frac{d x}{x\left(x^4+1\right)}=$

If $\int \frac{2x+3}{x(x+1)(x+2)(x+3)+1} dx = \frac{-1}{ax^2+bx+c} + \alpha$,then the value of $a+b+c$ is equal to

Integrate the function: $\frac{\cos x}{\sqrt{4-\sin ^{2} x}}$

$\int \frac{\log \sqrt{x}}{3 x} d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module