0
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...
Wiki User
∙ 13y ago
Add your answer:
Earn +20 pts
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
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
Integral of sin squared x?
First we look at the double-angle identity of cos2x. We know that: cos2x = cos^2x - sin^2x cos2x = [1-sin^2x] - sin^2x.............. (From sin^2x + cos^2x = 1, cos^2x = 1 - sin^2x) Therefore: cos2x = 1 - 2sin^2x 2sin^2x = 1 - cos2x sin^2x = 1/2(1-cos2x) sin^2x = 1/2 - cos2x/2 And intergrating, we get: x/2 - sin2x/4 + c...................(Integral of cos2x = 1/2sin2x; and c is a constant)
Find the derivative of y x2 sin x 2xcos x - 2sin x?
y = (x^2)(sin x)(2x)(cos x) - 2sin xy' = [[(x^2)(sin x)][(2x)(cos x)]]' - (2sin x)'y' = [[(x^2)(sin x)]'[(2x)(cos x)] + [(2x)(cos x)]'[(x^2)(sin x)]]- (2sin x)'y' = [[(x^2)'(sin x) + (sin x)'(x^2)][(2x)(cos x)] + [(2x)'(cos x) + (cos x)'(2x)][(x^2)(sin x)] ] - 2(cos x)y' = [[(2x)(sin x )+ (cos x)(x^2)][(2x)(cos x)] + [2cos x - (sin x)(2x)][(x^2)(sin x)]] - 2(cos x)y' = (4x^2)(sin x cos x) + (2x^3)(cos x)^2 + (2x^2)(sin x cos x) - (2x^3)(sin x)^2 - 2cos xy' = (6x^2)(sin x cos x) + (2x^3)(cos x)^2 - (2x^3)(sin x)^2 - 2cos x (if you want, you can stop here, or you can continue)y' = (3x^2)(2sin x cos x) + (2x^3)[(cos x)^2 - (sin x)^2] - 2cos xy' = (3x^2)(sin 2x) + (2x^3)(cos 2x) - 2 cos xy' = (2x^3)(cos 2x) + (3x^2)(sin 2x) - 2 cos x
Integration of sin2x?
Integral( sin(2x)dx) = -(cos(2x)/2) + C
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
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 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
Integral of sin squared x?
First we look at the double-angle identity of cos2x. We know that: cos2x = cos^2x - sin^2x cos2x = [1-sin^2x] - sin^2x.............. (From sin^2x + cos^2x = 1, cos^2x = 1 - sin^2x) Therefore: cos2x = 1 - 2sin^2x 2sin^2x = 1 - cos2x sin^2x = 1/2(1-cos2x) sin^2x = 1/2 - cos2x/2 And intergrating, we get: x/2 - sin2x/4 + c...................(Integral of cos2x = 1/2sin2x; and c is a constant)
How you can defferentiate an integral?
If the upper limit is a function of x and the lower limit is a constant, you can differentiate an integral using the Fudamental Theorem of Calculus. For example you can integrate Integral of [1,x^2] sin(t) dt as: sin(x^2) d/dx (x^2) = sin(x^2) (2x) = 2x sin(x^2) The lower limit of integration is 1 ( a constant). The upper limit of integration is a function of x, here x^2. The function being integrated is sin(t)
What is the integral of cos raised to 2 times x?
∫cos2(x).dxUse the identity cos2(x) = (1/2)(1+cos(2x))∫(1/2)(1+cos(2x))dxPull out constant:(1/2)∫(1+cos(2x))dxIntegrate:(1/2)(x + sin(2x)/2) + CSimplify:x/2 + sin(2x)/4 + CThe identity sin(2x) = 2sin(x)cos(x) can be used to rewrite it as(x + sin(x)cos(x))/2 + C
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
What is the indefinte integral of sin2x?
The indefinite integral of sin 2x is -cos 2x / 2 + C, where C is any constant.
Find the derivative of y x2 sin x 2xcos x - 2sin x?
y = (x^2)(sin x)(2x)(cos x) - 2sin xy' = [[(x^2)(sin x)][(2x)(cos x)]]' - (2sin x)'y' = [[(x^2)(sin x)]'[(2x)(cos x)] + [(2x)(cos x)]'[(x^2)(sin x)]]- (2sin x)'y' = [[(x^2)'(sin x) + (sin x)'(x^2)][(2x)(cos x)] + [(2x)'(cos x) + (cos x)'(2x)][(x^2)(sin x)] ] - 2(cos x)y' = [[(2x)(sin x )+ (cos x)(x^2)][(2x)(cos x)] + [2cos x - (sin x)(2x)][(x^2)(sin x)]] - 2(cos x)y' = (4x^2)(sin x cos x) + (2x^3)(cos x)^2 + (2x^2)(sin x cos x) - (2x^3)(sin x)^2 - 2cos xy' = (6x^2)(sin x cos x) + (2x^3)(cos x)^2 - (2x^3)(sin x)^2 - 2cos x (if you want, you can stop here, or you can continue)y' = (3x^2)(2sin x cos x) + (2x^3)[(cos x)^2 - (sin x)^2] - 2cos xy' = (3x^2)(sin 2x) + (2x^3)(cos 2x) - 2 cos xy' = (2x^3)(cos 2x) + (3x^2)(sin 2x) - 2 cos x
Integration of sin2x?
Integral( sin(2x)dx) = -(cos(2x)/2) + C
What is the integral of sine of x square?
It seems you can't express it in terms of the standard functions used in basic calculus; the site Wolfram Alpha (input: integral sin x^2) lists the integral in terms of a so-called Fresnel function. It also lists the first terms of the infinite series.
Why sin integral is cos?
sin integral is -cos This is so because the derivative of cos x = -sin x
