Question

If $y=\int \frac{dx}{{(1+x^2)}^{3/2}}$ and $y=0$ when $x=0$, then value of y when $x=1$, is

Answer 1

We are given ,

$y=\int {\frac{dx}{(1+x^2}}^{3/2}$$=\int \frac{dx}{(1+x{)}^2(1+x^2{)}^{1/2}}$

Let $x=tan\theta$

$dx=se{c}^2\theta d\theta$ ( diffentitation wrt x )

$\therefore dx=(1+ta{n}^2\theta )d\theta$

$dx=(1+x^2)d\theta$

$\therefore \frac{dx}{1+x^2}=d\theta$

$so,y=\int \frac{d\theta }{(1+ta{n}^2\theta {)}^{1/2}}$

$y=\int \frac{d\theta }{\sqrt{se{c}^2\theta }}$

$y=\int \frac{d\theta }{|sec\theta |}=\int |cos\theta |d\theta$

$y=sin\theta +C$

now , $sin\theta =tan\theta \times \frac{sin\theta }{tan\theta }=tan\theta .cos\theta$

$so,sin\theta =\frac{tan\theta }{\sqrt{se{c}^2\theta }}=\frac{tan\theta }{\sqrt{1+ta{n}^2\theta }}=\frac{x}{\sqrt{1+x^2}}$

now , at x = 0 we get y =0

$so,O=\frac{O}{\sqrt{1+0}}+C\to C=0$

$so,y=\frac{x}{\sqrt{1+x^2}}$

now at $x=1\to y=\frac{1}{\sqrt{1+1}}=\frac{1}{\sqrt{2}}$

the value of of y is $\frac{1}{\sqrt{2}}$ at $x=1$