Integral tanxsec^2x dx/√a^2+b^2tan^2x

1.	Integral tanxsec^2x dx/√a^2+b^2tan^2x
Answer» <U>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 :- $\boxed{\displaystyle{\sf{=\frac{\sqrt{b^2tan^2(x)a^2} }{b^2} +C}}}\:\:\:\:\:\:\:\:\:\:\:\: \mathbf{\cdots ANSWER}$

Answer»

<U>GIVEN EXPRESSION :-

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