Evaluate the definite integrals: [tex]\( \int_{0}^{\pi} (1 + \sin t + \cos t) \, dt + \int_{\pi}^{2\pi} (1 + \sin t + \cos t) \, dt \)[/tex]

To evaluate the definite integral ∫ 0 2 π ​ ( 1 + sin t + cos t ) d t , we can break it into two separate integrals as given:

∫ 0 π ​ ( 1 + sin t + cos t ) d t

∫ π 2 π ​ ( 1 + sin t + cos t ) d t


Let's evaluate each one step-by-step:
Step 1: Evaluate ∫ 0 π ​ ( 1 + sin t + cos t ) d t
Firstly, we integrate each term separately:

The integral of 1 with respect to t is t .

The integral of sin t is − cos t .

The integral of cos t is sin t .


So, the integral becomes: ∫ 0 π ​ ( 1 + sin t + cos t ) d t = [ t − cos t + sin t ] 0 π ​
Evaluating this from 0 to π :
( π − ( − 1 ) + 0 ) − ( 0 − 1 + 0 ) = ( π + 1 ) − ( − 1 ) = π + 2
Step 2: Evaluate ∫ π 2 π ​ ( 1 + sin t + cos t ) d t
We perform similar integration for this range: ∫ π 2 π ​ ( 1 + sin t + cos t ) d t = [ t − cos t + sin t ] π 2 π ​
Evaluating from π to 2 π :
( 2 π − ( − 1 ) + 0 ) − ( π − ( − 1 ) + 0 ) = ( 2 π + 1 ) − ( π + 1 ) = π
Final Result: Combine Both Integrals
Add the two results together: π + 2 + π = 2 π + 2
So, the value of the definite integral ∫ 0 2 π ​ ( 1 + sin t + cos t ) d t is 2 π + 2 .

Answered by DanielJosephParker | 2025-07-07