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Question

Evaluate the definite integral $$\displaystyle \int_{\tfrac {\pi}{6}}^{\tfrac {\pi}{4}}co\sec x dx$$


Solution

Let $$\displaystyle I=\int_{\tfrac {\pi}{6}}^{\tfrac {\pi}{4}}co\sec x dx$$
$$\Rightarrow\displaystyle  \int co\sec x dx=\log |co\sec x-\cot x|=F(x)$$
By second fundamental theorem of calculus, we obtain
$$\displaystyle I=F\left (\cfrac {\pi}{4}\right )-F\left (\cfrac {\pi}{6}\right )$$
$$\displaystyle =\log |co\sec \cfrac {\pi}{4}-\cot \cfrac {\pi}{4}|-\log |co\sec \cfrac {\pi}{6}-\cot \cfrac {\pi}{6}|$$
$$\displaystyle =\log|\sqrt 2-1|-\log|2-\sqrt 3|$$
$$\displaystyle =\log \left (\cfrac {\sqrt 2-1}{2-\sqrt 3}\right )$$

Mathematics
RS Agarwal
Standard XII

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