Mathematics

$$\displaystyle \int \sqrt{1-\sin x} \ dx$$


SOLUTION
$$I= \displaystyle\int \sqrt{1-\sin x}dx$$

We have $$\sqrt{1-\sin x}=\sqrt{\sin^{2}\dfrac{x}{2}+\cos^{2}\dfrac{x}{2}-2 \sin\dfrac{x}{2}\cos\dfrac{x}{2}}$$

$$=\sqrt{\left(\sin\dfrac{x}{2}-\cos\dfrac{x}{2}\right)^{2}}$$

$$=\sin\dfrac{x}{2}-\cos\dfrac{x}{2}$$

$$\therefore I= \displaystyle\int \left(\sin\dfrac{x}{2}-\cos\dfrac{x}{2}\right)dx$$

$$= \displaystyle\int \sin\dfrac{x}{2}dx-\int \cos\dfrac{x}{2}dx$$

$$=-\dfrac{\cos\dfrac{x}{2}}{\dfrac{1}{2}}-\dfrac{\sin\dfrac{x}{2}}{\dfrac{1}{2}}=-2\left[\cos\dfrac{x}{2}+\sin\dfrac{x}{2}\right]$$
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Subjective Medium Published on 17th 09, 2020
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