Question
Find the integrals of the functions in Exercises:
$\sin^{-1}(\cos\text{x})$
$\sin^{-1}(\cos\text{x})$
Then,$\ \sin\text{x}=\sqrt{1-\text{t}^2}$
$\Rightarrow(-\sin\text{x})\text{dx}=\text{dt}$ $\text{dx}=\frac{-\text{dt}}{\sin\text{x}}$ $\text{dx}=\frac{-\text{dt}}{\sqrt{1-\text{t}^2}}$ $\therefore\int\sin^{-1}(\cos\text{x})\text{dx}=\int\sin^{-1}\text{t}\Bigg(\frac{-\text{dt}}{\sqrt{1+\text{t}^2}}\Bigg)$ $=-\int\frac{\sin^{-1}\text{t}}{\sqrt{1-\text{t}^2}}\text{dt}$ $\text{Let }\sin^{-1}\text{t}=\text{u}$ $\Rightarrow\frac{1}{\sqrt{1-\text{t}^2}}\text{dt}=\text{du}$ $\therefore\int\sin^{-1}(\cos\text{x})\text{dx}=\int\text{u}\text{ du}$ $=-\frac{\text{u}^2}{2}+\text{C}$ $=\frac{-(\sin^{-1}\text{t})^2}{2}+\text{C}$ $=\frac{-\big[\sin^{-1}(\cos\text{x})\big]^2}{2}+\text{C} \ \ \ \ \ ...\text{(1)}$ It is known that, $\sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}$ $\therefore\sin^{-1}(\cos\text{x})=\frac{\pi}{2}-\cos^{-1}(\cos\text{x})=\bigg(\frac{\pi}{2}-\text{x}\bigg)$ Substituting in equation(1),we obtain $\int\sin^{-1}(\cos\text{x})\text{ dx}=\frac{-\bigg[\frac{\pi}{2}-\text{x}\bigg]^2}{2}+\text{C}$ $=-\frac{1}{2}\bigg(\frac{\pi^2}{2}+\text{x}^2-\pi\text{x}\bigg)+\text{C}$ $=-\frac{\pi^2}{8}-\frac{\text{x}^2}{2}+\frac{1}{2}\pi\text{x}+\text{C}$ $=\frac{\pi\text{x}}{2}-\frac{\text{x}^2}{2}+\bigg (\text{C}-\frac{\pi^2}{8}\bigg)$ $=\frac{\pi\text{x}}{2}-\frac{\text{x}^2}{2}+\text{C}$
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