Question

Evaluate the definite integral: $∫^\frac{π}{4}_\frac{π}{6} cosec x dx$

Answer 1

Let I=$∫^\frac{π}{4}_\frac{π}{6} cosec x dx$.

$∫cosec x dx=log|cosec x-cot x|=F(x)$

By second fundamental theorem of calculus,we obtain

$I=F(\frac{π}{4})-F(\frac{π}{6})$

$=log|cosec π/4-cot \frac{π}{4}|-log|cosec\frac{π}{6}-cot\frac{π}{6|}$

$=log|√2-1|-log|2-√3|$

$=log(\frac{√2-1}{2-√3})$