Answer on Question #50927 – Math – Integral Calculus
Find the ∫tan^3 xsec^3 xdx
A tan2x+1
B cot2x+1
C sec2x+1
D sec2x
Solution
𝑰 = ∫ 𝒕𝒂𝒏𝟑 𝒙 ∙ 𝒔𝒆𝒄𝟑 𝒙𝒅𝒙 =
= ∫ 𝒕𝒂𝒏𝒙 ∙ 𝒕𝒂𝒏𝟐 𝒙 ∙ 𝒔𝒆𝒄𝟑 𝒙𝒅𝒙 = ∫(𝒕𝒂𝒏𝒙 ∙ 𝒔𝒆𝒄𝟐 𝒙 − 𝒕𝒂𝒏𝒙)𝒔𝒆𝒄𝟑 𝒙𝒅𝒙
=
= ∫(𝐬𝐞𝐜 𝟓 𝒙 ∙ 𝒕𝒂𝒏𝒙 − 𝒔𝒆𝒄𝟑 𝒙 ∙ 𝒕𝒂𝒏𝒙)𝒅𝒙 = ∫(𝒔𝒆𝒄𝟒 𝒙 − 𝒔𝒆𝒄𝟐 𝒙)𝒔𝒆𝒄𝒙 ∙ 𝒕𝒂𝒏𝒙𝒅𝒙 =
= {𝒖 = 𝒔𝒆𝒄𝒙 =
= ∫(𝒖𝟒 − 𝒖𝟐 )𝒅𝒖 =
𝟏
,
𝒄𝒐𝒔𝒙
𝒅𝒖 = −
−𝒔𝒊𝒏𝒙𝒅𝒙
= 𝒕𝒂𝒏𝒙 ∙ 𝒔𝒆𝒄𝒙𝒅𝒙} =
𝒄𝒐𝒔𝟐 𝒙
𝟏 𝟓 𝟏 𝟑
𝟏
𝟏
𝒖 − 𝒖 + 𝑪 = 𝒔𝒆𝒄𝟓 𝒙 − 𝒔𝒆𝒄𝟑 𝒙 + 𝑪,
𝟓
𝟑
𝟓
𝟑
where 𝑪 is an arbitrary real constant.
www.AssignmentExpert.com