Bài 101146

Question

Tính tích phân \(I = \int\limits_0^\frac{\pi }{2}} {\left( {cos^{10}x + \sin^{10}}x - \cos^4x.\sin^4x\right)dx\)

hoàng anh thọ · Answer 1 · 4/25/2012 5:02:42 PM

\(f\left( x \right) = c{\rm{o}}{{\rm{s}}^{10}}x + {\sin ^{10}}x - c{\rm{o}}{{\rm{s}}^4}x.{\sin ^4}x = c{\rm{o}}{{\rm{s}}^{10}}x + {\sin ^{10}}x - c{\rm{o}}{{\rm{s}}^4}x.{\sin ^4}x\left( {c{\rm{o}}{{\rm{s}}^2}x + {{\sin }^2}x} \right)\)
\( = \left( {c{\rm{o}}{{\rm{s}}^4}x - {{\sin }^4}x} \right)\left( {c{\rm{o}}{{\rm{s}}^6}x - {{\sin }^6}x} \right)\)
\( = \left( {c{\rm{o}}{{\rm{s}}^2}x - {{\sin }^2}x} \right)^2\left( {c{\rm{o}}{{\rm{s}}^4}x + {{\sin }^4}x} +\sin^2 x\cos^2 x\right)\)
\( = \frac{1}{4}\cos^22x\left( {c{\rm{o}}{{\rm{s}}^4}x + {{\sin }^4}x} +\sin^2 x\cos^2 x\right)\)
\( = \frac{1}{4}\cos^22x\left( 1-\frac{1}{4}\sin^22 x\right)\)
\( = \frac{{15}}{{32}} + \frac{1}{2}c{\rm{os}}4x + \frac{1}{{32}}c{\rm{os}}8x\)
\(\Rightarrow I = \int\limits_0^{\frac{\pi }{2}} {f\left( x \right)} dx = \frac{{15}}{{32}}.\frac{\pi }{2} + \frac{1}{8}\sin 4x\left| {_0^{\pi /2}} \right. + \frac{1}{{32.8}}\sin 8x\left| {_0^{\pi /2}} \right. \Rightarrow I = \frac{{15\pi }}{{64}}\)

Bài 101146

1 Đáp án

Lý thuyết liên quan

Bài 101146

1 Đáp án

Liên quan

Lý thuyết liên quan