Evaluate the definite integral.?

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Answer 1 · 2017-11-16T20:21:19

$\frac{2 \sqrt{3} - 3}{3}$ $(2sqrt3-3)/3$

$\int_{0}^{\frac{π}{6}} \frac{sin (t)}{{cos}^{2} (t)} d t$ $int_0^(pi/6)sin(t)/cos^2(t)dt$

$= \int_{0}^{\frac{π}{6}} \frac{sin (t)}{cos (t)} \cdot \frac{1}{cos (t)} d t$ $=int_0^(pi/6)sin(t)/cos(t)*1/cos(t)dt$

$= \int_{0}^{\frac{π}{6}} tan (t) sec (t) d t$ $=int_0^(pi/6)tan(t)sec(t)dt$

$= {[sec (t)]}_{0}^{\frac{π}{6}}$ $=[sec(t)]_0^(pi/6)$

$= sec (\frac{π}{6}) - sec (0)$ $=sec(pi/6)-sec(0)$

$= \frac{2}{\sqrt{3}} - 1$ $=2/sqrt3-1$

$= \frac{2 - \sqrt{3}}{\sqrt{3}}$ $=(2-sqrt3)/sqrt3$

$= \frac{2 \sqrt{3} - 3}{3}$ $=(2sqrt3-3)/3$

Note that $\frac{d}{d t} sec (t) = sec (t) tan (t)$ $d/dtsec(t)=sec(t)tan(t)$ (used above)