Integral (COS x) / (3 + sin square x) DX

Integral (COS x) / (3 + sin square x) DX

=Integral 1 / (3 + sin squared x) d (SiN x)
=Integral 1 / (3 + x ^ 2) DX
Then there is the basic formula. Check the book

Integral of ∫ e ^ SiN x / (e ^ SiN x + e ^ cos x) DX on 0 ~ π / 2

Let x = π / 2 - t, DX = - DT when x = 0, t = π / 2, when x = π / 2, t = 0l = ∫ (0 -- > π / 2) e ^ SiNx / (e ^ SiNx + e ^ cosx) DX = ∫ (π / 2 -- > 0) e ^ sin (π / 2 - t) / [e ^ sin (π / 2 - t) + e ^ cos (π / 2 - t)] · - DT = ∫ (0 -- > π / 2) e ^ cost / (e ^

∫sin^3(x)cos^2(x)dx=

Take out a sin (x)
∫sin^3(x)cos^2(x)dx
=-∫sin^2(x)cos^2(x)d(cos(x))
=-∫(1-cos^2)cos^2(x)d(cos(x))
=-∫cos^2-cos^4(x)d(cos(x))
=-(cos^3(x))/3+(cos^5(x))/5+c

F (x) = 2Sin ^ 2 (π / 4 + x) + radical 3 (sin ^ 2x + cos ^ 2x) F (x) = 2Sin ^ 2 (π / 4 + x) + radical 3 (sin ^ 2x cos ^ 2x) symbol is wrong

f(x)=2sin^2(π/4+x)+√3(sin^2x-cos^2x)
=2sin^2(π/4+x)-√3cos2x
=1-cos(π/2+2x)+√3cos2x
=1+2(√3/2cos2x+1/2sin2x)
=1+2sin(π/3+2x)

Does it exist α ∈(-π/2,π/2), β ∈ (0, π), so that the equation sin (3 π- α)= Radical 2 cos (π) / 2- β), Root 3cos (- α）=- Root 2Sin Connect 7 π / 2- β） Simultaneously established? If it exists, find αβ The value of does not exist, indicating the reason,

Using the induction formula to simplify the two known equations by sin (3 π- α)= √2cos(π/2- β) Get sin α= √2sin β ① By √ 3cos (- α）=- √2sin(7π/2- β) Get √ 3cos α= √2cos β ② Add the squares of the two formulas to get sin ²α+ 3COS ²α= 2 convert cosine into sine to get sin ²α...

Calculus, ∫ radical (x ^ 2 + A ^ 2) DX For example, find the indefinite integral, ∫ root sign (x ^ 2 + A ^ 2) DX,

Let x = atan θ, Change the root sign into ASEC θ, Then the partial integral is used, and the result is 0.5A ^ 2 (SEC) θ tan θ+ ln/sec θ+ tan θ/)+ C. Then put x = atan θ Just take it back