Evaluate the integral? : # int_0^(pi/6) sinx/cos^2x dx #

Question

Evaluate the integral? : # int_0^(pi/6)  sinx/cos^2x dx #

Emilia Aguilar · Accepted Answer

# int_0^(pi/6) \ sinx/cos^2x \ dx = 2/sqrt(3) - 1 #

We seek: # I = int_0^(pi/6) \ sinx/cos^2x \ dx # We can perform a simple substitution: Let #u=cosx => (du)/dx = -sinx # And we must change the limits of integration: When # x= { (0), (pi/6) :} => u = { (1), (sqrt(3)/2) :}# So substituting into the integral, we get: # I = int_0^(pi/6) \ (-1)/cos^2x \ (-sinx) \ dx # # \ \ = - int_1^(sqrt(3)/2) \ 1/u^2 \ du # # \ \ = [1/u]_1^(sqrt(3)/2) # # \ \ = 1/(sqrt(3)/2) - 1/1# # \ \ = 2/sqrt(3) - 1 #

Emilia Aguilar · Answer

undefined To evaluate the integral $\int_0^{\frac{\pi}{6}} \frac{\sin(x)}{\cos^2(x)} \, dx$, we can use the substitution method. Let $u = \cos(x)$, then $du = -\sin(x) \, dx$.

When $x = 0$, $u = \cos(0) = 1$ and when $x = \frac{\pi}{6}$, $u = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$.

Now, we rewrite the integral in terms of $u$:

$$\int_{1}^{\frac{\sqrt{3}}{2}} \frac{-1}{u^2} \, du$$

$$= \left[-\frac{1}{-1}\right]_{1}^{\frac{\sqrt{3}}{2}}$$

$$= \left[1 - \frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2}\right]$$

$$= 1 - \frac{4}{3}$$

$$= \frac{3}{3} - \frac{4}{3}$$

$$= \frac{-1}{3}$$

Therefore, $\int_0^{\frac{\pi}{6}} \frac{\sin(x)}{\cos^2(x)} \, dx = \frac{-1}{3}$.