Question

plz answer this question guys​

Answer 1

Given :

$\displaystyle{\sf \int \dfrac{dx}{4 sin^2x+5cos^2x}}$

Solution :

$\Rightarrow\displaystyle{\sf \int \dfrac{dx}{4 sin^2x+5cos^2x}}$

$\Rightarrow \displaystyle {\sf \int \dfrac{sec^2x}{4tan^2x+5}dx}$

$\boxed{\sf x = t \Rightarrow sec^2xdx=dt}$

$\displaystyle{\implies\sf \int \dfrac{dt}{4t^2+5}}$

$\displaystyle{\implies \sf \dfrac{1}{4} \int \dfrac{dt}{t^2+\Big(\dfrac{5}{4}\Big)}}$

${\displaystyle{\implies \sf \dfrac{1}{4} \times \dfrac{1}{\dfrac{\sqrt{5}}{2}}\:tan^{-1}\Bigg(\dfrac{t}{\dfrac{\sqrt{5}}{2}}\Bigg)+C}}$

${\displaystyle{\implies \sf \dfrac{1}{2\sqrt{5}}\:tan^{-1}\Big(\dfrac{2t}{\sqrt{5}}\Big)+C}}$

${\displaystyle{\implies \sf \dfrac{1}{2\sqrt{5}}\:tan^{-1}\Big(\dfrac{2 tanx}{\sqrt{5}}\Big)+C}}$

Required Answer :

$\boxed{\boxed{\displaystyle{\sf \int \dfrac{dx}{4 sin^2x+5cos^2x}}={\displaystyle{ \sf \dfrac{1}{2\sqrt{5}}\:tan^{-1}\Big(\dfrac{2 tanx}{\sqrt{5}}\Big)+C}}}}$

Answer 2

Answer:

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