Question

integarte dx/2sin^2x+5cos^2x​

Answer 1

$\underline{\textbf{Given:}}$

$\displaystyle\mathsf{\int\;\dfrac{1}{2\,sin^2x+5\,cos^2x}dx}$

$\underline{\textbf{To find:}}$

$\displaystyle\mathsf{\int\;\dfrac{1}{2\,sin^2x+5\,cos^2x}dx}$

$\underline{\textbf{Solution:}}$

$\underline{\textbf{Formula used:}}$

$\boxed{\displaystyle\mathsf{\int\;\dfrac{1}{x^2+a^2}dx=tan^{-1}\left(\dfrac{x}{a}\right)+C}}$

$\mathsf{Consider,}$

$\displaystyle\mathsf{\int\;\dfrac{1}{2\,sin^2x+5\,cos^2x}dx}$

$\textsf{This can be written as}$

$\displaystyle\mathsf{=\int\;\dfrac{1}{cos^2x\left(2\,\dfrac{sin^2x}{cos^2x}+5\right)}dx}$

$\displaystyle\mathsf{=\int\;\dfrac{sec^2x}{2\,tan^2x+5}dx}$

$\textsf{We apply the following substituition}$

$\boxed{\mathsf{Take\;t=tanx\;\implies\;\dfrac{dt}{dx}=sec^2x\;\implies\;dt=sec^2x\,dx}}$

$\displaystyle\mathsf{=\int\;\dfrac{1}{2\,t^2+5}dt}$

$\displaystyle\mathsf{=\dfrac{1}{2}\int\;\dfrac{1}{t^2+\dfrac{5}{2}}dt}$

$\displaystyle\mathsf{=\dfrac{1}{2}\int\;\dfrac{1}{t^2+\left(\dfrac{\sqrt{5}}{\sqrt{2}}\right)^2}dt}$

$\mathsf{=\dfrac{1}{2}{\times}\dfrac{1}{\dfrac{\sqrt{5}}{\sqrt{2}}}\,tan^{-1}\left(\dfrac{t}{\dfrac{\sqrt{5}}{\sqrt{2}}}\right)+C}$

$\mathsf{=\dfrac{1}{\sqrt{2}}{\times}\dfrac{1}{\sqrt{5}}\,tan^{-1}\left(\dfrac{\sqrt{2}t}{\sqrt{5}}\right)+C}$

$\mathsf{=\dfrac{1}{\sqrt{10}}\,tan^{-1}\left(\dfrac{\sqrt{2}tanx}{\sqrt{5}}\right)+C}$

$\underline{\textbf{Find more:}}$

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