高数积分∫1/（cos⁴x+sin⁴x）dx

高数积分∫1/（cos⁴x+sin⁴x）dx
高数积分∫1/（cos⁴x+sin⁴x）dx

高数积分∫1/（cos⁴x+sin⁴x）dx
您好

∫1/（cos⁴x+sin⁴x）dx
=∫1/[(cos^2x+ sin^2 x)^2-2(sinxcosx)^2]dx
=∫1/（1-1/2sin²2x）dx
=∫2/（2-sin²2x）dx
=∫2/（1+cos²2x）dx
=2∫sec²2x/（sec²2x+1）dx
=∫1/（tan²2x+2)d(tan2x)
=1/√2*arctan(tan2x/√2)+C
=√2/2*arctan(tan2x/√2)+C

【数学辅导团】为您解答,不理解请追问,理解请及时选为满意回答!(*^__^*)谢谢!

原式=∫dx/[(cos^2x+ sin^2 x)^2-2(sinxcosx)^2]
=∫dx/[1-0.5(sin2x)^2]
令u=tan2x, 则x=(arctanu)/2
dx=0.5du/(1+u^2)
原式=∫0.5udu/(1+0.5u^2)
=0.5∫d(0.5u^2)/(1+0.5u^2)
=0.5 ln(1+0.5u^2)+C

全部展开

原式=∫dx/[(cos^2x+ sin^2 x)^2-2(sinxcosx)^2]
=∫dx/[1-0.5(sin2x)^2]
令u=tan2x, 则x=(arctanu)/2
dx=0.5du/(1+u^2)
原式=∫0.5udu/(1+0.5u^2)
=0.5∫d(0.5u^2)/(1+0.5u^2)
=0.5 ln(1+0.5u^2)+C
=0.5 ln[1+0.5(tan2x)^2]+C

收起