Question

$\int \frac{se{c}^{2}x}{\sqrt{ase{c}^{2}x-bta{n}^{2}x}}dx$ is $($where $c$ is integration constant$)$

• A. $\frac{1}{\sqrt{b-a}}si{n}^{-1}\left(tanx\sqrt{\frac{b-a}{a}}\right)+c$ if $b>a>0$.
• B. $\frac{1}{\sqrt{b-a}}lo{g}_2(tanx\sqrt{b-a}+\sqrt{ase{c}^{2}x-bta{n}^{2}x})+c$ if $b>a>0$
• C. $\frac{1}{\sqrt{a-b}}si{n}^{-1}\left(tanx\sqrt{\frac{a-b}{a}}\right)+c$ if $a>b>0$.
• D. $\frac{1}{\sqrt{a-b}}lo{g}_e(tanx\sqrt{a-b}+\sqrt{ase{c}^{2}x-bta{n}^{2}x})+c$ if $a>b>0$

Answer 1

Answer: (A), (D)

$\frac{1}{\sqrt{b-a}}si{n}^{-1}\left(tanx\sqrt{\frac{b-a}{a}}\right)+c$ if $b>a>0$.
$\frac{1}{\sqrt{a-b}}lo{g}_e(tanx\sqrt{a-b}+\sqrt{ase{c}^{2}x-bta{n}^{2}x})+c$ if $a>b>0$

$I=\int \frac{se{c}^{2}x}{\sqrt{ase{c}^{2}x-bta{n}^{2}x}}dx$
$=\int \frac{se{c}^{2}x}{\sqrt{a(1+ta{n}^{2}x)-bta{n}^{2}x}}dx$
put $tanx=t$
$\Rightarrow se{c}^{2}xdx=dt$
$\Rightarrow I=\int \frac{dt}{\sqrt{a+t^2(a-b)}}$
Since we know,
$\int \frac{dx}{\sqrt{a^2-x^2}}=si{n}^{-1}\left(\frac{x}{a}\right)+c$

$\int \frac{dx}{\sqrt{a^2+x^2}}=lo{g}_e(x+\sqrt{x^2+a^2})+c$
Now If $a>b>0$
$I=\int \frac{dt}{\sqrt{{\left(\sqrt{a}\right)}^2+{\left(t\sqrt{a-b}\right)}^2}}$

$=\frac{1}{\sqrt{a-b}}lo{g}_e(t\sqrt{a-b}+\sqrt{a+t^2(a-b)})$

$=\frac{1}{\sqrt{a-b}}lo{g}_e(t\sqrt{a-b}+\sqrt{a+se{c}^{2}x-bta{n}^{2}x})+c$
and If $b>a>0$

$I=\int \frac{dt}{\sqrt{{\left(\sqrt{a}\right)}^2-{\left(t\sqrt{b-a}\right)}^2}}=\frac{1}{\sqrt{b-a}}si{n}^{-1}\left(\frac{t\sqrt{b-a}}{\sqrt{a}}\right)+c$

$=\frac{1}{\sqrt{b-a}}si{n}^{-1}\left(tanx\sqrt{\frac{b-a}{a}}\right)+c$