Question

integrate the question in the attached image

Answer 1

Answer:

$\text{Let }u=e^x$

$\text{Differentiating both sides with respect to }x,\\\\\dfrac{du}{dx}=\dfrac{d(e^x)}{dx}=e^x\\\\\text{or, }du=e^xdx$

$\therefore\ \int{e^x\sqrt{1-e^{2x}}\ dx}=\int \sqrt{1-u^2}\ du$

$\text{Again, let }u=\text{sin}\theta \Rightarrow \boxed{\theta=\text{sin}^{-1}u}\\\\\text{Differentiating both sides with respect to }\theta,\\\\\dfrac{du}{d\theta}=\dfrac{d(\text{sin}\theta)}{d\theta}\\\\\text{or, }\dfrac{du}{d\theta}=\text{cos}\theta\\\\\text{or, }du=\text{cos}\theta\ d\theta$

$\therefore\ \int \sqrt{1-u^2}\ du=\int\sqrt{1-\text{sin}^2\theta}\ \text{cos}\theta d\theta=\int \text{cos}\theta\ \text{cos}\theta\ d\theta=\int \text{cos}^2\theta d\theta$

$=\int\dfrac{\text{cos}2\theta+1}{2}d\theta\\\\=\dfrac{1}{2}\bigl[\int \text{cos}2\theta d\theta +\int1 d\theta\bigl]\\\\=\dfrac{1}{2}\biggl[\dfrac{\text{sin}2\theta}{2}+\theta\biggl]+c$

$=\dfrac{1}{4}\text{sin}2\theta+\dfrac{1}{2}\theta+c\\\\=\dfrac{1}{4}(2\text{sin}\theta\text{cos}\theta)+\dfrac{1}{2}\text{sin}^{-1}u+c\\\\=\dfrac{1}{2}\text{sin}\theta\sqrt{1-\text{sin}^2\theta}+\dfrac{1}{2}\text{sin}^{-1}e^x+c$

$=\dfrac{1}{2}u\sqrt{1-u^2}+\dfrac{1}{2}\text{sin}^{-1}e^x+c$

$\therefore\ \int e^x\sqrt{1-e^{2x}}\ dx=\dfrac{1}{2}e^x\sqrt{1-e^{2x}}+\dfrac{1}{2}\text{sin}^{-1}e^x+c$