Let's ask a question about calculus. For example, for sin (x), the integral should be - cos (x). Suppose that the upper and lower limits of the definite integral of sin (x) DX are 2 and 1. Then it should be equal to (- cos (2)) - (- cos (1)), but the two calculated by calculator are not the same. Is it necessary to convert 2 and 1 into f (a) - f (b), where a and B are the upper and lower limits of sin (x) integral, and F is the original function of integrand What should I do in writing?

Let's ask a question about calculus. For example, for sin (x), the integral should be - cos (x). Suppose that the upper and lower limits of the definite integral of sin (x) DX are 2 and 1. Then it should be equal to (- cos (2)) - (- cos (1)), but the two calculated by calculator are not the same. Is it necessary to convert 2 and 1 into f (a) - f (b), where a and B are the upper and lower limits of sin (x) integral, and F is the original function of integrand What should I do in writing?

The main problem is that the integral is limited to radians, and the mode of the calculator should be changed to "rad" for calculation

Solving ∫ 1 / (COS ^ 4 (x) sin ^ 2 (x)) DX

∫1/[(cosx)^4(sinx)]dx=∫[(sinx)^2+(cosx)^2]/[(cosx)^4(sinx)]dx
=∫(secx)^4dx+4∫(csc2x)^2dx
∫(secx)^4dx=∫(secx)^2[(tanx)^2+1]dx=∫[(tanx)^2+1]dtanx=(tanx)^3/3+tanx
∫(csc2x)^2dx=-1/2*cot2x
So ∫ 1 / [(cosx) ^ 4 (SiNx)] DX = (TaNx) ^ 3 / 3 + tanx-2cot2x + C

∫ sin^4(x)cos^3(x)dx

∫ sin ^ 4 (x) cos ^ 3 (x) DX = ∫ sin ^ 4 (x) cos ^ 2 (x) cosx DX = ∫ sin ^ 4 (x) * [1 - Sin ^ 2 (x)] d (SiNx) = ∫ sin ^ 4 (x) d (SiNx) -∫ sin ^ 6 (x) d (SiNx) = (1 / 5) * sin ^ 5 (x) - (1 / 7) sin ^ 7 (x) + constant

Solving integral ∫ (1 / (SiN x cos x)) DX

If the numerator and denominator are multiplied by cosx, the numerator will become dsinx, and the denominator (cosx) &# 178; = 1 - (SiNx) &# 178;, and the variable SiNx can be replaced by T

∫ 1/((4(cos x)^2)-((sin x)^2)) dx
How to solve this problem?

Solution: divide the numerator and denominator by (COS x) ^ 2 at the same time, and then use the first substitution integral method (i.e. approximate differential method) to deal with: ∫ 1 / [4 (COS x) ^ 2 - (SiN x) ^ 2] DX = ∫ (secx) ^ 2 / [4 - (Tan x) ^ 2] DX = ∫ 1 / [4 - (Tan x) ^ 2] d (TaNx) = 1 / 4 ∫ [1 / (2 + Tan x

By the method of substitution: ∫ 1 / (2 (SiN x) ^ 2 + 3 (COS x) ^ 2) DX

Let t = TaNx, then DT = D (TaNx)
The original formula = ∫ DX / (2Sin & sup2; X + 3cos & sup2; x)
=∫dx/[cos²x(2sin²x/cos²x+3)]
=∫sec²xdx/(2sin²x/cos²x+3)
=∫d(tanx)/(2tan²x+3)
=∫dt/(2t²+3)
=1/3∫dt/(2t²/3+1)
=1/√6∫d(√(2/3)t)/[(√(2/3)t)²+1]
=1 / √ 6arctan [√ (2 / 3) TaNx] + C (C is an integral constant)

1 ∫ cos & sup2; x-sin & sup2; X / cos & sup2; xsin & sup2; X DX 2 ∫ 1 + cos & sup2; X / 1 + cos2x DX
1 2 separate

It's a pity that I graduated more than 4 years ago. Please refuel yourself. I'm sorry

Finding the indefinite integral of (1 + x ^ 2) cos (NX) DX
Such as the title, will not count, hope to give the detailed process, there are additional!

(1+x^2)cos(nx)dx∫▒〖(1+x^2)cos(nx)dx〗1/n ∫▒〖cos(nx)dx+〗  1/n ∫▒x^2  con(nx)dx1/n  sin⁡(nx)+1/n ∫▒x^2  con(nx)dx1/n  sin⁡(nx)+1/n  sin⁡(nx)*x^2-2/n^2  ∫▒x sin(nx)dx1/n  sin⁡ (nx)+1/n  sin⁡(nx)*x^2+2/n^2  xcon(nx)-2/n^2  ∫▒〖con(nx)〗 dx1/n  sin⁡(nx)+1/n  sin⁡(nx)*x^2+2/n^2  xcon(nx)-2/n^3  sin⁡(nx)

How to find the indefinite integral of ∫ X & # 178; sinxdx ∫ cos (2x-1) DX by the method of partial integration

∫x²sinxdx
u=x² 2x 2 0
v'=sinx -cosx -sinx cosx
∫x²sinxdx=-x²cosx+2xsinx+2cosx+c
∫cos﹙2x-1﹚dx
=1/2∫cos﹙2x-1﹚d(2x-1)
=1/2sin(2x-1)+c

How to integrate ∫ 1 / cos (2x) DX
Such as the title
Can you give me a solution? My answer to ∫ 1 / cos (NX) DX is (1 / N) (LN | cos (NX / 2) - sin (NX / 2) | - ln | sin (NX / 2) + cos (NX / 2) |) + C. how can this be solved?

1/cos(nx) = 1/(cos²(nx/2) - sin²(nx/2))
= (1/2)((cos(nx/2) - sin(nx/2))/(cos(nx/2) + sin(nx/2)) - (-cos(nx/2) - sin(nx/2))/(cos(nx/2) - sin(nx/2)))
∫(cos(nx/2) - sin(nx/2))/(cos(nx/2) + sin(nx/2))dx
=(2 / N) ∫ (1 / T) DT (let (COS (NX / 2) + sin (NX / 2) = t)
=(2/n)ln|t|
=(2/n)ln|cos(nx/2) + sin(nx/2)| + C1
In the same way
∫(-cos(nx/2) - sin(nx/2))/(cos(nx/2) - sin(nx/2))dx
=(2/n)ln|cos(nx/2) - sin(nx/2)| + C2
∫1/cos(nx)dx = ∫1/(cos²(nx/2) - sin²(nx/2))dx
= ∫(1/2)((cos(nx/2) - sin(nx/2))/(cos(nx/2) + sin(nx/2)) - (-cos(nx/2) - sin(nx/2))/(cos(nx/2) - sin(nx/2)))dx
=(1/2)((2/n)ln|cos(nx/2) + sin(nx/2)| - (2/n)ln|cos(nx/2) - sin(nx/2)|) + C
=(1/n)(ln|cos(nx/2) + sin(nx/2)| - ln|cos(nx/2) - sin(nx/2)|) + C
Can you give me a solution? The answer to ∫ 1 / cos (NX) DX is (1 / N) (ln| cos (NX / 2) - sin (NX / 2) | - ln| sin (NX / 2) + cos (NX / 2) | + C. how can this be solved?
Just verify it,
(1/n)(ln|cos(nx/2)-sin(nx/2)|-ln|sin(nx/2)+cos(nx/2)|)+c
Derivation = (1 / N) (n / 2)
((-sin(nx/2)-cos(nx/2))/(cos(nx/2)-sin(nx/2))
- (cos(nx/2)-sin(nx/2))/(sin(nx/2)+cos(nx/2)))
=(1/2)(-2)/(cos²(nx/2) - sin²(nx/2))
= -1/cos(nx)
So the answer is wrong