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The question is ambiguous: does it refer to 1/sin(x) + cos(x) or to 1/[sin(x)+cos(x)]?

The question is ambiguous: does it refer to 1/sin(x) + cos(x) or to 1/[sin(x)+cos(x)]?

The question is ambiguous: does it refer to 1/sin(x) + cos(x) or to 1/[sin(x)+cos(x)]?

The question is ambiguous: does it refer to 1/sin(x) + cos(x) or to 1/[sin(x)+cos(x)]?

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โˆ™ 2013-03-06 21:43:39
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โˆ™ 2013-03-06 21:43:39

The question is ambiguous: does it refer to 1/sin(x) + cos(x) or to 1/[sin(x)+cos(x)]?

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Q: What is the integral of 1 divided by sin x plus cos x?
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Why sin integral is cos?

sin integral is -cos This is so because the derivative of cos x = -sin x


Integral of 1 divided by sinx cosx?

Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C


Verify that sin minus cos plus 1 divided by sin plus cos subtract 1 equals sin plus 1 divided by cos?

[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,


What is the integral of cosx divided by sinx plus cosx from 0 to 2pi?

The Integral diverges. It has singularities whenever sin(x)+cos(x)=0. Singularities do not necessarily imply that the integral goes to infinity, but that is the case here, since the indefinite integral is x/2 + 1/2 Log[-Cos[x] - Sin[x]]. Obviously this diverges when evaluated at zero and 2pi.


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 would you solve the integral of 1 plus tan2x plus tan squared 2x?

Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C


Integral of x cosx dx?

The integral of x cos(x) dx is cos(x) + x sin(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


Factor sin cubed plus cos cubed?

sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)


What is the integral of sin x?

-cos x + Constant


Integration of cos 5x?

The integral of cos 5x is 1/5 sin (5x)


How do you simplify cos times cot plus sin?

cos*cot + sin = cos*cos/sin + sin = cos2/sin + sin = (cos2 + sin2)/sin = 1/sin = cosec


Cos x sin x integral?

sin2x + c


How do you show that 2 sin squared x minus 1 divided by sin x minus cos x equals sin x plus cos x?

(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x


Integral of sin x sin 2x?

Ok, I know that sin 2x can be substituted out for 2 sin x cos x. so now I have the Integral of (sin x ) ( 2 sin x cos x ) dx which is 2 sin2x cos x dx If I use integration by parts with the u and dv, I find myself right back again..can you help. The book gives an answer, but I am not sure how it was achieved. the book gives 2/3 sin 2x cos x - 1/3 cos 2x sin x + C I may be making this more difficult than it is ? when you get the integral of 2 sin2x cos x dx use u substitution. u= sinx du= cosxdx. Then you'll get the integral of 2u^2 du.. .and then integrate...


What is the derivative of cos x sin x divided by cos x sin x?

(cos x sin x) / (cos x sin x) = 1. The derivative of a constant, such as 1, is zero.


Integral of tan square x secant x?

convert tan^2x into sin^2x/cos^2x and secant x into 1/cos x combine terms for integral sin^2x/cos^3x dx then sub in u= cos^3x and du=-2sin^2x dx


Int sin 3x cos 5x dx?

integral sin(3 x) cos(5 x) dx = 1/16 (8 cos^2(x)-cos(8 x))+C


What is the integral for sin 1-x?

- cos(1 - X) + C


What is the integral of -5.4 sin kt?

(5.4 / k) cos(kt)


What is integral square root of sin x.dx?

-cos(x) + constant


Why does sin plus cos equal one?

There is no reason at all. For most angles sin plus cos do not equal one.


Integral of sec2x-cosx plus x2dx?

I wasn't entirely sure what you meant, but if the problem was to find the integral of [sec(2x)-cos(x)+x^2]dx, then in order to get the answer you must follow a couple of steps:First you should separate the problem into three parts as you are allowed to with integration. So it becomes the integral of sec(2x) - the integral of cos(x) + the integral of x^2Then solve each part separatelyThe integral of sec(2x) is -(cos(2x)/2)The integral of cos(x) is sin(x)The integral of x^2 isLastly you must combine them together:-(cos(2x)/2) - sin(x) + (x^3)/3


What is the value of sin A plus cos Acos A (1-cos A) when tan A 815?

When tan A = 815, sin A = 0.9999992 and cos A = 0.0012270 so that sin A + cos A*cos A*(1-cos A) = 1.00000075, approx.


What is the integral of sin x Times Square x times exp x?

Assume the expression is: ∫ sin(x)x²e^x dx Then: Take the integral: integral e^x x^2 sin(x) dx For the integrand e^x x^2 sin(x), integrate by parts, integral f dg = f g- integral g df, where f = x^2, dg = e^x sin(x) dx, df = 2 x dx, g = 1/2 e^x (sin(x)-cos(x)): = 1/2 e^x x^2 sin(x)-1/2 (e^x x^2 cos(x))- integral e^x x (sin(x)-cos(x)) dx Expanding the integrand e^x x (sin(x)-cos(x)) gives e^x x sin(x)-e^x x cos(x): = 1/2 e^x x^2 sin(x)-1/2 (e^x x^2 cos(x))- integral (e^x x sin(x)-e^x x cos(x)) dx Integrate the sum term by term and factor out constants: = 1/2 e^x x^2 sin(x)-1/2 (e^x x^2 cos(x))- integral e^x x sin(x) dx+ integral e^x x cos(x) dx For the integrand e^x x sin(x), integrate by parts, integral f dg = f g- integral g df, where f = x, dg = e^x sin(x) dx, df = dx, g = 1/2 e^x (sin(x)-cos(x)): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x x sin(x)+1/2 e^x x cos(x)+ integral e^x x cos(x) dx+1/2 integral e^x (sin(x)-cos(x)) dx Expanding the integrand e^x (sin(x)-cos(x)) gives e^x sin(x)-e^x cos(x): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x x sin(x)+1/2 e^x x cos(x)+ integral e^x x cos(x) dx+1/2 integral (e^x sin(x)-e^x cos(x)) dx Integrate the sum term by term and factor out constants: = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x x sin(x)+1/2 e^x x cos(x)+1/2 integral e^x sin(x) dx-1/2 integral e^x cos(x) dx+ integral e^x x cos(x) dx For the integrand e^x cos(x), use the formula integral exp(alpha x) cos(beta x) dx = (exp(alpha x) (alpha cos(beta x)+beta sin(beta x)))/(alpha^2+beta^2): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/4 e^x sin(x)-1/2 e^x x sin(x)-1/4 (e^x cos(x))+1/2 e^x x cos(x)+1/2 integral e^x sin(x) dx+ integral e^x x cos(x) dx For the integrand e^x sin(x), use the formula integral exp(alpha x) sin(beta x) dx = (exp(alpha x) (alpha sin(beta x)-beta cos(beta x)))/(alpha^2+beta^2): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x x sin(x)-1/2 (e^x cos(x))+1/2 e^x x cos(x)+ integral e^x x cos(x) dx For the integrand e^x x cos(x), integrate by parts, integral f dg = f g- integral g df, where f = x, dg = e^x cos(x) dx, df = dx, g = 1/2 e^x (sin(x)+cos(x)): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)+e^x x cos(x)-1/2 e^x cos(x)-1/2 integral e^x (sin(x)+cos(x)) dx Expanding the integrand e^x (sin(x)+cos(x)) gives e^x sin(x)+e^x cos(x): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)+e^x x cos(x)-1/2 e^x cos(x)-1/2 integral (e^x sin(x)+e^x cos(x)) dx Integrate the sum term by term: = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)+e^x x cos(x)-1/2 e^x cos(x)-1/2 integral e^x sin(x) dx-1/2 integral e^x cos(x) dx For the integrand e^x cos(x), use the formula integral exp(alpha x) cos(beta x) dx = (exp(alpha x) (alpha cos(beta x)+beta sin(beta x)))/(alpha^2+beta^2): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/4 e^x sin(x)+e^x x cos(x)+-3/4 e^x cos(x)-1/2 integral e^x sin(x) dx For the integrand e^x sin(x), use the formula integral exp(alpha x) sin(beta x) dx = (exp(alpha x) (alpha sin(beta x)-beta cos(beta x)))/(alpha^2+beta^2): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x sin(x)+e^x x cos(x)-1/2 e^x cos(x)+constant Which is equal to: Answer: | | = 1/2 e^x ((x^2-1) sin(x)-(x-1)^2 cos(x))+constant