Solve: $\frac{π}{3} \int \frac{π}{6} cos x d x$

Open in App

Solution

Now,
$\frac{π}{3} \int \frac{π}{6} cos x d x$

$= {[sin x]}_{\frac{π}{6}}^{\frac{π}{3}}$

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

$= \frac{\sqrt{3} - 1}{2}$ .

Suggest Corrections

Similar questions

\int_{0}^{2 π} x I n (\frac{3 + c o s x}{3 - c o s x}) d x =

___

Q. Solve

π \int \frac{π}{2} \frac{1 - sin x}{1 - cos x} d x

Q. Solve:

{lim}_{x \to \frac{π}{6}} \frac{sin (x - \frac{π}{6})}{\sqrt{3} - 2 cos x}

Solve the equation: $\sqrt{(\frac{1}{16} + c o s^{4} x - \frac{1}{2} c o s^{2} x)} + \sqrt{(\frac{9}{16} + c o s^{4} x - \frac{3}{2} c o s^{2} x)} = \frac{1}{2}$

\int_{0}^{π} \frac{tan x}{sec x + cos x} d x =

Join BYJU'S Learning Program

Solve: π3∫π6cosxdx

Now,π3∫π6cosxdx=[sinx]π3π6=√32−12=√3−12.

Solve: $\frac{π}{3} \int \frac{π}{6} cos x d x$

Now,
$\frac{π}{3} \int \frac{π}{6} cos x d x$

$= {[sin x]}_{\frac{π}{6}}^{\frac{π}{3}}$

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

$= \frac{\sqrt{3} - 1}{2}$ .