∫ ( tan^2 x + tan^4 x )dx
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∫(tan^2x+tan^4x)dx∫(tan^2x+tan^4x)dx∫(tan^2x+tan^4x)dx∫(tan²x+tan⁴x)dx=∫tan²x(1+tan&
∫ ( tan^2 x + tan^4 x )dx
∫ ( tan^2 x + tan^4 x )dx
∫ ( tan^2 x + tan^4 x )dx
∫ ( tan²x + tan⁴x) dx
= ∫ tan²x (1 + tan²x) dx
= ∫ tan²x sec²x dx
= ∫ tan²x d tanx
= (1/3) tan³x + c
∫ ( tan^2 x + tan^4 x )dx
=∫ ((secx)^2-1 +((secx)^2-1) ^2 )dx
=∫ ( (secx)^2-1+(secx)^4-2*(secx)^2+1)dx
=∫ ((secx)^4-( secx)^2)dx
∫( ( secx)^2-1)dtanx
=∫ ( tanx )^2dtanx
=1/3*(tanx)^3+c
∫ ( tan^2 x + tan^4 x )dx
求不定积分∫(tan^2x+tan^4x)dx
tan（ x/2+π/4）+tan（x/2-π/4 ）=2tanxtan(x/2+π/4)+tan(x/2-π/4)=[tan(x/2)+tan(π/4)]/[1-tan(x/2)tan(π/4)]+[tan(x/2)-tan(π/4)]/[1+tan(x/2)tan(π/4)]=[tan(x/2)+1]/[1-tan(x/2)]+[tan(x/2)-1]/[1+tan(x/2)]=[(tan(x/2)+1)^2-(tan(x/2)-1)^2]/[1-(tan(x
tan(20)+2*tan(40)+4*tan(10)-tan(70)
不定积分(tan^2x+tan^4x)dx
这道题为什么第二步的+到了第三步变成了-1.tan(x/2+π/4)+tan(x/2-π/4)=[tan(x/2)+tan(π/4)]/[1-tan(x/2)tan(π/4)]+[tan(x/2)-tan(π/4)]/[1+tan(x/2)tan(π/4)]=[tan(x/2)+1]/[1-tan(x/2)]+[tan(x/2)-1]/[1+tan(x/2)]=[(tan(x/2)+1)^2-(tan(x/2)-
tan(-x+π/4)
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