Best solutions to - "Evaluate the integrals ∫0π2[cos⁡ti−sin⁡2tj+sin2tk]dt"

midtlinjeg

Answered question

2021-10-15

Evaluate the integrals

\int_{0}^{\frac{π}{2}} [\cos t i - \sin 2 t j + \sin^{2} t k] d t

sovienesY

Skilled2021-10-16Added 89 answers

Step 1: Given that
Evaluate the integrals

\int_{0}^{\frac{π}{2}} [\cos t i - \sin 2 t j + \sin^{2} t k] d t

Step 2: Solve
We have,

\int_{0}^{\frac{π}{2}} [\cos t \hat{i} - \sin 2 t \hat{j} + \sin^{2} t \hat{k}] d t = \int_{0}^{\frac{π}{2}} \cos t \hat{i} d t - \int_{0}^{\frac{π}{2}} \sin 2 t \hat{j} d t + \int_{0}^{\frac{π}{2}} \sin^{2} t \hat{k} d t

= \hat{i} {[\sin t]}_{0}^{\frac{π}{2}} + \frac{\hat{j}}{2} {[\cos 2 t]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} (\frac{1 - \cos 2 t}{2}) \hat{k} d t

= \hat{i} {[\sin t]}_{0}^{\frac{π}{2}} + \frac{\hat{j}}{2} {[\cos 2 t]}_{0}^{\frac{π}{2}} + \frac{\hat{k}}{2} {[t - \frac{\sin 2 t}{2}]}_{0}^{\frac{π}{2}}

= \hat{i} [\sin (\frac{π}{2}) - \sin (0)] + \frac{\hat{j}}{2} [\cos (2 \times \frac{π}{2}) - \cos (0)] + \frac{\hat{k}}{2} [\frac{π}{2} - 0 - \frac{\sin 2 (\frac{π}{2})}{2} + 0]

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