Solve: $\int_{- π / 2}^{π / 2} {sin}^{2} x d x$

Open in App

Solution

$\int_{- π / 2}^{π / 2} \frac{1 - cos 2 x}{2} d x$
$= {[\frac{x}{2} - \frac{sin 2 x}{4}]}_{- π / 2}^{π / 2}$
$= \frac{π}{2}$

Suggest Corrections

Similar questions

\int_{- π / 2}^{π / 2} \sin^{2} x d x .

\int_{- \frac{π}{2}}^{\frac{π}{2}} s i n^{2} x d x =

\int_{- \frac{π}{2}}^{\frac{π}{2}} {sin}^{2} x d x

By using properties of definite integrals, evaluate the integrals
Let $\int_{- \frac{π}{2}}^{\frac{π}{2}} s i n^{2} x d x .$

\int_{- π / 2}^{π / 2} cos x d x

Join BYJU'S Learning Program