Misc 27 - Definite integral cos2 x dx / cos2 x + 4 sin2 x

Misc 27 Evaluate the definite integral ﷐0﷮﷐𝜋﷮2﷯﷮﷐﷐﷐cos﷮2﷯﷮𝑥﷯ 𝑑𝑥﷮﷐﷐cos﷮2﷯﷮𝑥﷯ + 4﷐﷐sin﷮2﷯﷮𝑥﷯﷯ ﷯ Let I = ﷐0﷮﷐𝜋﷮2﷯﷮﷐﷐𝑐𝑜𝑠﷮2﷯𝑥﷮﷐𝑐𝑜𝑠﷮2﷯𝑥 + 4﷐𝑠𝑖𝑛﷮2﷯𝑥﷯﷯ 𝑑𝑥 = ﷐0﷮﷐𝜋﷮2﷯﷮﷐﷐𝑐𝑜𝑠﷮2﷯𝑥﷮﷐𝑐𝑜𝑠﷮2﷯𝑥 + 4 ﷐﷐1 − 𝑐𝑜𝑠﷮2﷯𝑥﷯ 𝑑𝑥﷯﷯ = ﷐0﷮﷐𝜋﷮2﷯﷮﷐﷐𝑐𝑜𝑠﷮2﷯𝑥﷮4 − 3 ﷐𝑐𝑜𝑠﷮2﷯𝑥﷯﷯ 𝑑𝑥 = ﷐−1﷮3﷯ ﷐0﷮﷐𝜋﷮4﷯﷮ ﷐−3 ﷐𝑐𝑜𝑠﷮2﷯ 𝑥 ﷮4 − 3 ﷐𝑐𝑜𝑠﷮2﷯𝑥﷯ ﷯ 𝑑𝑥 = −﷐1﷮3﷯﷐0﷮﷐𝜋﷮2﷯﷮﷐﷐4 −3 𝑐𝑜𝑠﷮2﷯𝑥 −4﷮4 − 3 ﷐𝑐𝑜𝑠﷮2﷯𝑥﷯﷯ 𝑑𝑥 = −﷐1﷮3﷯﷐0﷮﷐𝜋﷮2﷯﷮1−﷐4﷮4 − 3 ﷐𝑐𝑜𝑠﷮2﷯𝑥﷯﷯ 𝑑𝑥 = −﷐1﷮3﷯﷐0﷮﷐𝜋﷮2﷯﷮𝑑𝑥+﷐4﷮3﷯﷐0﷮﷐𝜋﷮2﷯﷮﷐𝑑𝑥﷮4 − 3 ﷐𝑐𝑜𝑠﷮2﷯𝑥﷯﷯﷯ Dividing numerator and denominator by ﷐𝑐𝑜𝑠﷮2﷯ 𝑥 = −﷐1﷮3﷯﷐﷐𝜋﷮2﷯﷯+﷐4﷮3﷯ ﷐0﷮﷐𝜋﷮2﷯﷮﷐﷐𝑑𝑥﷮﷐𝑐𝑜𝑠﷮2﷯ 𝑥﷯﷮﷐4 − 3 ﷐𝑐𝑜𝑠﷮2﷯𝑥 ﷮﷐𝑐𝑜𝑠﷮2﷯ 𝑥﷯﷯﷯ 𝑑𝑥 = −﷐1﷮3﷯﷐﷐𝜋﷮2﷯﷯+﷐4﷮3﷯ ﷐0﷮﷐𝜋﷮2﷯﷮﷐﷐𝑠𝑒𝑐﷮2﷯𝑥﷮4 ﷐𝑠𝑒𝑐﷮2﷯𝑥 − 3﷯﷯ 𝑑𝑥 = −﷐𝜋﷮6﷯+﷐4﷮3﷯ ﷐0﷮﷐𝜋﷮2﷯﷮﷐﷐𝑠𝑒𝑐﷮2﷯𝑥﷮4 (1 + ﷐𝑡𝑎𝑛﷮2﷯𝑥) − 3﷯﷯ 𝑑𝑥 Put tan x = t so that ﷐𝑠𝑒𝑐﷮2﷯ x dx = dt Thus, When x = 0, t = 0, and when x = ﷐𝜋﷮2﷯, 𝑡= 𕔴 Substituting values and limit ∴ I =−﷐𝜋﷮6﷯+﷐4﷮3﷯ ﷐0﷮𕔴uc1﷮﷐𝑑𝑡﷮4 ﷐1+ ﷐𝑡﷮2﷯﷯ −3﷯﷯ =−﷐𝜋﷮6﷯+﷐4﷮3﷯ ﷐0﷮𕔴uc1﷮﷐𝑑𝑡﷮﷐4𝑡﷮2﷯+1﷯﷯ =−﷐𝜋﷮6﷯+﷐4﷮3﷯ ∙﷐1﷮4﷯ ﷐0﷮𕔴uc1﷮﷐𝑑𝑡﷮﷐ 𝑡﷮2﷯+﷐1﷮4﷯﷯﷯ =﷐𝜋﷮6﷯+﷐4﷮3﷯∙﷐1﷮4﷯ × ﷐1﷮﷐﷐1﷮2﷯﷯﷯ ﷐﷐﷐𝑡𝑎𝑛﷮−1﷯﷐𝑡﷮﷐1﷮2﷯﷯﷯﷮0﷮𕔴uc1﷯ = −﷐𝜋﷮6﷯+﷐2﷮3﷯∙ ﷐﷐﷐𝑡𝑎𝑛﷮−1﷯ 2𝑡﷯﷮0﷮𕔴uc1﷯ =−﷐𝜋﷮6﷯+﷐2﷮3﷯∙﷐[𝑡𝑎𝑛﷮−1﷯𕔴uc1−﷐𝑡𝑎𝑛﷮−1﷯0] = −﷐𝜋﷮6﷯+﷐2﷮3﷯∙﷐﷐𝜋﷮2﷯−0﷯ =−﷐𝜋﷮6﷯+﷐𝜋﷮3﷯ =﷐𝝅﷮𝟔﷯

Misc 27 - Chapter 7 Class 12 Integrals