Mathematics

Evaluate
$$\int \sin^{-1} \dfrac{2x}{1+x^2} dx$$


SOLUTION
$$I=\int \sin^{-1}\dfrac{2x}{1+x^{2}}dx$$

Put $$x=\tan t$$

Then, $$sin^{-1} \dfrac{2x}{1+x^{2}}=sin^{-1} \dfrac{2\tan t}{1+tan^{2}t}$$

$$=sin^{-1} sin2t$$

$$=2t$$

Thus, our function becomes

$$I=2\int tan^{-1}xdx$$

$$I=2tan^{-1}x \int 1. dx -2\int (\dfrac{d(tan^{-1}x)}{dx} \int 1.dx)dx$$

$$I=2xtan^{-1}x-2\int \dfrac{1}{1+x^{2}}xdx$$

Let $$I_{1}=\int \dfrac{x}{1+x^{2}}dx$$

Let $$1+x^{2}=t$$

$$dx=\dfrac{dt}{2x}$$

Therefore,

$$I_{1}=\int \dfrac{x}{t}. \dfrac{dt}{2x}$$

$$I_{1}=\dfrac{1}{2}\int \dfrac{dt}{t}$$

$$I_{1}=\dfrac{1}{2}\log t$$

$$I_{1}=\dfrac{1}{2}\log (1+x^{2})$$

Hence,

$$I=2xtan^{-1}x-\log (1+x^{2})+C$$
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Subjective Medium Published on 17th 09, 2020
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