Question

integration of sin square wt dt limits 0 to T

Answer 1

that's it.......................

Answer 2

Answer:

The integration of given expression is \frac{1}{2} [\((T-\frac{sin2wT}{2w})

Step-by-step explanation:

Required is the integration of below expression:

$\int\limits^T_0 {sin^{2}wt } \, dt$

we know that $sin^{2}x=\frac{1-cos2x}{2}$, put this value in above integration equation, we get below:

$\int\limits^T_0 {\frac{1-cos2wt}{2}  } \, dt \\\frac{1}{2} \int\limits^T_0 {1-cos2wt} \, dt$

integrating above expression, we get

$\frac{1}{2} \int\limits^T_0 {1-cos2wt} \, dt \\\frac{1}{2} \((t-\frac{sin2wt}{2w})$put value of given limits

$\frac{1}{2} [\((T-\frac{sin2wT}{2w})-[\((0-\frac{sin2w(0)}{2w})]\\\frac{1}{2} [\((T-\frac{sin2wT}{2w})$

hence the integration of given expression is \frac{1}{2} [\((T-\frac{sin2wT}{2w})