Mathematics

Evaluate:

$$\displaystyle\int {\dfrac{1}{\sqrt{1-(bx+c)^2}}dx}$$


SOLUTION
$$I=\displaystyle\int \dfrac{1}{\sqrt{1-(bx+c)^2}}dx$$

put $$x=\left[\dfrac{\cos\theta}{b}-c\right]$$

$$\Rightarrow \cos\theta =bx+c$$

$$\theta =\cos^{-1}(bx+c)$$

$$dx=\dfrac{-\sin\theta}{b}d\theta$$

$$I=\displaystyle\int \dfrac{-\sin\theta d\theta}{b\sqrt{1-\cos^2\theta}}$$

$$I=\displaystyle\int \dfrac{-\sin\theta}{b\sin\theta}d\theta$$

$$I=-\dfrac{1}{b}\displaystyle\int d\theta =-\dfrac{1}{b}\theta +k$$

$$\therefore  \displaystyle\int \dfrac{1}{\sqrt{1-(bx+c)^2}}=-\dfrac{1}{b}\cos^{-1}(bx+c)+k$$.
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Subjective Medium Published on 17th 09, 2020
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