Evaluate the integrals \int_{0}^{\frac{\pi}{2}}[\cos t i - \sin 2t j

midtlinjeg

midtlinjeg

Answered question

2021-10-15

Evaluate the integrals
0π2[costisin2tj+sin2tk]dt

Answer & Explanation

sovienesY

sovienesY

Skilled2021-10-16Added 89 answers

Step 1: Given that
Evaluate the integrals
0π2[costisin2tj+sin2tk]dt
Step 2: Solve
We have,
0π2[costi^sin2tj^+sin2tk^]dt=0π2costi^dt0π2sin2tj^dt+0π2sin2tk^dt
=i^[sint]0π2+j^2[cos2t]0π2+0π2(1cos2t2)k^dt
=i^[sint]0π2+j^2[cos2t]0π2+k^2[tsin2t2]0π2
=i^[sin(π2)sin(0)]+j^2[cos(2×π2)cos(0)]+k^2[π20sin2(π2)2+0]

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?