$\int {sin}^{2} b x d x$

Open in App

Solution

$\int {sin}^{2} b x d x$

$= \frac{1}{2} \int 2 {sin}^{2} b x d x$

$= \frac{1}{2} \int (1 - cos 2 b x) d x$

$= \frac{1}{2} [x - \frac{sin 2 b x}{2 b}] + c$

$= \frac{x}{2} - \frac{sin 2 b x}{4 b} + c$

Suggest Corrections

Similar questions

\int (\frac{x}{a + b x}) d x =

\int \frac{1}{a^{x} b^{x}} d x

\int \frac{1}{a^{x} b^{x}} d x

\int_{- 1}^{1} (a x^{3} + b x) d x = 0

\int \frac{1}{a^{x} b^{x}} d x

Join BYJU'S Learning Program