Mathematics

$$\displaystyle\int \dfrac{dx}{\sqrt{10-10x-2x^2}}$$=?


SOLUTION
We are given $$I=\displaystyle\int \dfrac{dx}{\sqrt{10-10x-2x^{2}}}$$
$$=\displaystyle\int \dfrac{dx}{\sqrt{\dfrac{45}{2}-(2x^{2}+10x+25/2)}}$$
$$=\displaystyle\dfrac{dx}{\sqrt{\left(\dfrac{3\sqrt{5}}{\sqrt{2}}\right)^{2}-\left(\sqrt{2}x+\dfrac{5}{\sqrt{2}}\right)^{2}}}$$
as we know, $$\displaystyle\int \dfrac{dx}{\sqrt{a^{2}-x^{2}}}=\sin^{-1}\left(\dfrac{x}{a}\right)+C$$
$$I=\sin^{-1}\left[\dfrac{\sqrt{2}x+5/\sqrt{2}}{\dfrac{3\sqrt{5}}{\sqrt{2}}}\right]+C$$
$$=\sin^{-1}\left(\dfrac{2x+5}{3\sqrt{5}}\right)+C$$
So, 
$$\displaystyle\int \dfrac{1}{\sqrt{10-10x-2x^{2}}}dx=\sin^{--1}\left(\dfrac{2x+5}{3\sqrt{5}}\right)+C$$
Where $$C$$ is any arbitary constant.
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Subjective Medium Published on 17th 09, 2020
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