Question

Integral tanxsec^2x dx/√a^2+b^2tan^2x

Answer 1

GIVEN EXPRESSION :-

$\sf{\dfrac{tan(x)sec^2(x)}{\sqrt{a^2+b^2tan^2(x)} }dx}$

OBJECTIVE :-

To integrate the given expression.

SOLUTION :-

$\displaystyle{\sf{\int \frac{sec^2(x)tan(x)}{\sqrt{b^2tan^2(x)+a^2} } dx}}$

$\rule{100}{1}$

$\displaystyle{\sf{Let\ b^2tan^2(x)+a^2=u}}\\\\\displaystyle{\sf{= Then\ \frac{du}{dx}=2b^2sec^2(x)tan(x)}\:\:\:\:\cdots(By\ differentiation)}$

$\displaystyle{\sf{\longrightarrow dx=\frac{1}{2b^2sec^2(x)tan(x)} du}}$

$\rule{100}{1}$

$\displaystyle{\sf{= \frac{1}{2b^2}\int \frac{1}{\sqrt{u} }du }}$

$\boxed{\sf{Solving\ \int \frac{1}{\sqrt{u}} du :-}}$

$\rule{100}1$

Using power rule :-

$\displaystyle{\sf{= \int u^\frac{-1}{2}du }}$

$\displaystyle{\sf{=\frac{u^{\frac{-1}{2} +1}}{\frac{-1}{2} +1} }}$

$\displaystyle{\sf{= 2\sqrt{u} }}$

$\rule{100}1$

$\displaystyle{\sf{\frac{1}{2b^2} \int \frac{1}{\sqrt{u} } du }}$

$\displaystyle{\sf{= \frac{\sqrt{u} }{b^2} }}$

We know, u = b²tan²(x) + a². Putting the value :-

$\boxed{\displaystyle{\sf{=\frac{\sqrt{b^2tan^2(x)a^2} }{b^2} +C}}}\:\:\:\:\:\:\:\:\:\:\:\: \mathbf{\cdots ANSWER}$