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Why not? Just a second integration. Drop the constant.
int[- cos(x)] dx
the negative implies - 1 and can be brought out side the integrand
- int[cos(x)] dx
= - sin(x) + C
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∙ 11y ago
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How do you integrate sinx divide sinx plus cosx?
1/2(x-ln(sin(x)+cos(x)))
Cos x equals -cos x plus 1?
No, but cos(-x) = cos(x), because the cosine function is an even function.
How do you prove that (1 plus cotx)2-2cotx 1(1-cos)(1 plus cos)?
Manipulate normally, noting:cot x = cos x / sin xcos² x + sin² x = 1 → sin²x = 1 - cos² xa² - b² = (a + b)(a - b)1 = 1²ab = baa/(bc) = a/b/c(1 + cot x)² - 2 cot x = 1² + 2 cot x + cot² x - 2 cot x= 1 + cot² x= 1 + (cos x / sin x)²= 1 + cos² x / sin² x= 1 + cos² x / (1 - cos² x)= ((1 - cos² x) + cos² x)/(1 - cos² x)= 1/(1² - cos² x)= 1/((1 + cos x)(1 - cos x))= 1/(1 - cos x)/(1 + cos x)QED.
How do you express sin x plus cos x divided by cos x in terms of tan x?
(sin x + cos x) / cosx = sin x / cos x + cosx / cos x = tan x + 1
How do you solve sinx divided by 1 plus cosx plus cosx divided by sinx?
sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
How do you integrate 2sinxcosx?
Integrate 2sin(x)cos(x)dxLet u = cos(x) and du = -sin(x)dx and pull out the -2:-2[Integral(u*du)]Integrate with respect to u:-2(u2)/2 + CSimplify:-u2 + CReplace u with cos(x):-cos2(x) + C
How do you integrate sinx divide sinx plus cosx?
1/2(x-ln(sin(x)+cos(x)))
What is the antiderivative of -cscxcotx?
First, antiderivative = a solution to the indefinite integral therefore to integrate -(csc(x))(cot(x)) first convert it to -cos(x)/sin2(x) To integrate ∫-cos(x)/sin2(x) dx, use substitution u = sin(x) and du/dx = cosx This will make it ∫-1/u2 du and the antiderivative is 1/u +c, therefore the answer is 1/sin(x) + c.
Integration of tangent x?
To integrate tan(x), you must break up tangent into sine over cosine, with that being done, all you have is a u-substitution with the cosine. This should give: int(tan(x)dx)=int(sin(x)/cos(x)dx)=int(-(1/u)*du)=-ln|u|+C=-ln|cos(x)|+C u=cos(x) du=-sin(x)dx
Integrate sin x cos 2x?
By Angle-Addition, cos(2x) = 2cos(x)^2-1 So, sin(x)cos(2x) = [2cos(x)^2-1]sin(x) = 2sin(x)cos(x)^2 - sin(x) Int[2sin(x)cos(x)^2 - sin(x)] = (-2/3)cos(x)^3 + cos(x) +K
Cos x equals -cos x plus 1?
No, but cos(-x) = cos(x), because the cosine function is an even function.
How do you prove that (1 plus cotx)2-2cotx 1(1-cos)(1 plus cos)?
Manipulate normally, noting:cot x = cos x / sin xcos² x + sin² x = 1 → sin²x = 1 - cos² xa² - b² = (a + b)(a - b)1 = 1²ab = baa/(bc) = a/b/c(1 + cot x)² - 2 cot x = 1² + 2 cot x + cot² x - 2 cot x= 1 + cot² x= 1 + (cos x / sin x)²= 1 + cos² x / sin² x= 1 + cos² x / (1 - cos² x)= ((1 - cos² x) + cos² x)/(1 - cos² x)= 1/(1² - cos² x)= 1/((1 + cos x)(1 - cos x))= 1/(1 - cos x)/(1 + cos x)QED.
What is the integration of tanx?
The integral of tan(x) dx = ln | sec(x) | + cto solve... tan(x) = sin(x)/cos(x)the integral of (sin(x)/cos(x) dx) ... let u = cos(x) then du = -sin(x) dx= the integral of (1/u -du)= -ln | u | + c= -ln | cos(x) | + c= ln | (cos(x))^-1 | + c ... or ... ln | 1/cos(x) | + c= ln | sec(x) | + c
How do you express sin x plus cos x divided by cos x in terms of tan x?
(sin x + cos x) / cosx = sin x / cos x + cosx / cos x = tan x + 1
How do you solve sinx divided by 1 plus cosx plus cosx divided by sinx?
sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
What is the integral of zero to pai by two x cos cubed x?
(3pi-7)/9 To verify this go to the link and enter integrate x cos(x)^3 from 0 to pi/2.
How do you differentiate cos squared x?
use the double angle formula for cos(2x) which is: cos(2x)=2cos^2(x)-1 by this relation cos^2(x)=(cos(2x)+1)/2 now we'd integrate this instead this will give sin(2x)/4+x/2 =) hope this helps
