Question 3a

SCH_CALC L3CALC

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Prove:

$$\frac{1}{1+\sin x}=\sec x(\sec x-\tan x)$$ $$RHS={\sec }^{2}x-\sec x\tan x$$ $$=\frac{1}{{\cos }^{2}x}-\frac{1}{\cos x}\cdot \frac{\sin x}{\cos x}$$ $$=\frac{1}{{\cos }^{2}x}-\frac{\sin x}{{\cos }^{2}x}$$ $$=\frac{1-\sin x}{{\cos }^{2}x}$$ $$=\frac{1-\sin x}{1-{\sin }^{2}x}$$ $$=\frac{1-\sin x}{(1+\sin x)(1-\sin x)}$$ $$=\frac{1}{1+\sin x}=LHS∎$$

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$${\int }_0^{\frac{2\pi }{3}}\frac{1}{1+\sin x}dx$$ $$={\int }_0^{2\pi /3}\sec x(\sec x-\tan x)dx$$ $$={\int }_0^{2\pi /3}{\sec }^{2}x-\sec x\tan xdx$$ $$={\left[\tan x-\sec x\right]}_0^{2\pi /3}$$ $$=\tan \left(\frac{2\pi }{3}\right)-\frac{1}{\cos \left(\frac{2\pi }{3}\right)}-\left(\tan 0-\frac{1}{\cos 0}\right)$$ $$=\boxed{3-\sqrt{3}}$$