Example 44

Transcript

Example 44 Evaluate 0﷮𝜋﷮ 𝑥 𝑑𝑥﷮ 𝑎﷮2﷯ cos﷮2﷯﷮𝑥 + 𝑏﷮2﷯﷯ sin﷮2﷯﷮𝑥﷯﷯﷯ Let 0﷮𝜋﷮ 𝑥﷮ 𝑎﷮2﷯𝑐𝑜 𝑠﷮2﷯𝑥 + 𝑏﷮2﷯𝑠𝑖 𝑛﷮2﷯𝑥﷯𝑑𝑥﷯ ∴ I= 0﷮𝜋﷮ 𝜋 − 𝑥﷯﷮ 𝑎﷮2﷯𝑐𝑜 𝑠﷮2﷯ 𝜋 − 𝑥﷯ + 𝑏﷮2﷯𝑠𝑖 𝑛﷮2﷯ 𝜋 − 𝑥﷯﷯𝑑𝑥﷯ I= 0﷮𝜋﷮ 𝜋 − 𝑥﷮ 𝑎﷮2﷯ 𝑐𝑜𝑠 𝜋 − 𝑥﷯﷯﷮2﷯ + 𝑏﷮2﷯ 𝑠𝑖𝑛 𝜋 − 𝑥﷯﷯﷮2﷯﷯𝑑𝑥﷯ I= 0﷮𝜋﷮ 𝜋 − 𝑥﷮ 𝑎﷮2﷯ − 𝑐𝑜𝑠 𝑥﷯﷮2﷯ + 𝑏﷮2﷯ 𝑠𝑖𝑛 𝑥﷯﷮2﷯﷯𝑑𝑥﷯ I= 0﷮𝜋﷮ 𝜋 − 𝑥﷮ 𝑎﷮2﷯ cos﷮2﷯﷮𝑥﷯ + 𝑏﷮2﷯ sin﷮2﷯﷮𝑥﷯﷯𝑑𝑥﷯ Adding (1) and (2) i.e. (1) + (2) I+I= 0﷮𝜋﷮ 𝑥﷮ 𝑎﷮2﷯ cos﷮2﷯﷮𝑥﷯ + 𝑏﷮2﷯ sin﷮2﷯﷮𝑥﷯﷯𝑑𝑥﷯+ ﷮﷮ 𝜋 − 𝑥﷮ 𝑎﷮2﷯ cos﷮2﷯﷮𝑥﷯ + 𝑏﷮2﷯ sin﷮2﷯﷮𝑥﷯﷯﷯𝑑𝑥 2I= 0﷮𝜋﷮ 𝑥 + 𝜋 − 𝑥﷮ 𝑎﷮2﷯ cos﷮2﷯﷮𝑥﷯ + 𝑏﷮2﷯ sin﷮2﷯﷮𝑥﷯﷯﷯𝑑𝑥 2I= 0﷮𝜋﷮ 𝜋 ﷮ 𝑎﷮2﷯ cos﷮2﷯﷮𝑥﷯ + 𝑏﷮2﷯ sin﷮2﷯﷮𝑥﷯﷯﷯𝑑𝑥 I= 𝜋﷮2﷯ 0﷮𝜋﷮ 1﷮ 𝑎﷮2﷯ cos﷮2﷯﷮𝑥﷯ + 𝑏﷮2﷯ sin﷮2﷯﷮𝑥﷯﷯𝑑𝑥﷯ Dividing numerator and denominator by 𝐶𝑜 𝑠﷮2﷯𝑥, we get I= 𝜋﷮2﷯ 0﷮𝜋﷮ 1﷮ cos﷮2﷯﷮𝑥﷯﷯﷮ 𝑎﷮2﷯ cos﷮2﷯﷮𝑥 + 𝑏﷮2﷯ sin﷮2﷯﷮𝑥﷯﷯﷮ cos﷮2﷯﷮𝑥﷯﷯ ﷯ 𝑑𝑥﷯ I= 𝜋﷮2﷯ 0﷮𝜋﷮ 𝑠𝑒 𝑐﷮2﷯𝑥﷮ 𝑎﷮2﷯ cos﷮2﷯﷮𝑥﷯﷮ cos﷮2﷯﷮𝑥﷯﷯ + 𝑏﷮2﷯ sin﷮2﷯﷮𝑥﷯﷮ cos﷮2﷯﷮𝑥﷯﷯﷯ 𝑑𝑥﷯ I= 𝜋﷮2﷯ 0﷮𝜋﷮ 𝑠𝑒 𝑐﷮2﷯𝑥﷮ 𝑎﷮2﷯ + 𝑏﷮2﷯ tan﷮2﷯﷮𝑥﷯﷯ 𝑑𝑥﷯ Let 𝑓 𝑥﷯= sec﷮2﷯﷮𝑥﷯﷮ 𝑎﷮2﷯+ 𝑏﷮2﷯ tan﷮2﷯﷮𝑥﷯﷯ 𝑓 2𝑎−𝑥﷯= sec﷮2﷯﷮ 𝜋 − 𝑥﷯﷯﷮ 𝑎﷮2﷯ + 𝑏﷮2﷯ tan﷮2﷯﷮ 𝜋 − 𝑥﷯﷯﷯ 𝑓 2𝑎−𝑥﷯= −𝑠𝑒𝑐 𝑥﷯﷮2﷯﷮ 𝑎﷮2﷯ + 𝑏﷮2﷯ − tan﷮𝑥﷯﷯﷮2﷯﷯ 𝑓 2𝑎−𝑥﷯= 𝑠𝑒 𝑐﷮2﷯𝑥﷮ 𝑎﷮2﷯ + 𝑏﷮2﷯ tan﷮2﷯﷮𝑥﷯﷯ ∴ 𝑓 𝑥﷯=𝑓 2𝑎−𝑥﷯ ∴ I= 𝜋﷮2﷯ 0﷮𝜋﷮ 𝑠𝑒 𝑐﷮2﷯𝑥﷮ 𝑎﷮2﷯ + 𝑏﷮2﷯ tan﷮2﷯﷮𝑥﷯﷯ 𝑑𝑥﷯= 𝜋﷮2﷯ × 2 0﷮ 𝜋﷮2﷯﷮ 𝑠𝑒 𝑐﷮2﷯𝑥﷮ 𝑎﷮2﷯ + 𝑏﷮2﷯ tan﷮2﷯﷮𝑥﷯﷯ 𝑑𝑥﷯ Let 𝑏 tan﷮𝑥=𝑡﷯ Differentiating both sides w.r.t.𝑥 𝑏 𝑠𝑒 𝑐﷮2﷯𝑥 𝑑𝑥=𝑑𝑡 𝑑𝑡= 𝑑𝑡﷮ 𝑏﷮2﷯ 𝑠𝑒 𝑐﷮2﷯ 𝑥﷯ Putting the values of tan 𝑥 and 𝑑𝑥 , we get 𝐼=𝜋 0﷮ 𝜋﷮2﷯﷮ 𝑠𝑒 𝑐﷮2﷯𝑥﷮ 𝑎﷮2﷯ + 𝑡﷮2﷯﷯ . 𝑑𝑥﷯ 𝐼=𝜋 0﷮唴﷮ 𝑠𝑒 𝑐﷮2﷯𝑥﷮ 𝑎﷮2﷯ + 𝑡﷮2﷯﷯ . 𝑑𝑡﷮𝑏 𝑠𝑒 𝑐﷮2﷯ 𝑥﷯﷯ 𝐼= 𝜋﷮𝑏﷯ 0﷮唴﷮ 𝑑𝑡﷮ 𝑎﷮2﷯ + 𝑡﷮2﷯﷯﷯ 𝐼= 𝜋﷮𝑏﷯ 1﷮𝑎﷯ tan﷮−1﷯﷮ 𝑡﷮𝑎﷯﷯﷯﷯﷮0﷮唴﷯ Putting limits, I= 𝜋﷮𝑏﷯ 1﷮𝑎﷯ 𝑡𝑎𝑛﷮−1﷯ 唴﷮𝑎﷯﷯− 1﷮𝑎﷯ 𝑡𝑎𝑛﷮−1﷯ 0﷮𝑎﷯﷯﷯ = 𝜋﷮𝑏﷯ 1﷮𝑎﷯ tan﷮−1﷯﷮ 唴﷯− 1﷮𝑎﷯ tan﷮−1﷯﷮ 0﷯﷯﷯﷯ = 𝜋﷮𝑏﷯ 1﷮𝑎﷯ 𝜋﷮2﷯﷯−0﷯ = 𝝅﷮𝟐﷯﷮𝟐𝒂𝒃﷯