1+cscx cotx integralf dx is ec X-CSCX

Q: 2 dydx + f(x, y)dydx

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Q: evaluate using Integration by Parts.

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Q: Find dy Y= 5In CI+x") 4 dy'=?

A: The solution are next step..

Q: Find integral.

A: Given: Let I=∫2x cos x2 dx To Find: To evaluate the given integration.

Q: Show that ∫0∞ e-x 2 dx = ∫10√(-ln y)  dy  by interpreting theintegrals as areas.

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Q: 6. Encuentre la solución general de la siguiente ecuación diferencial dy 2.ry(y – 1) dx 3(1 + a)

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Q: sen z + cos x dz | (tag z + 1) dx = In sec x+z + C 37. %3D cos 2

A: 37.   Given integral, ∫sen x+ cos xcos xdx

Q: dy 3. 1+ sin 4y

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Q: Evaluate the double integral : (3 – 8z) dz dy

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Q: Evaluate the integral ∫dx / 1+cscx.

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Q: tan- Find -dx using integration by parts. x2

A: The given integral is: I=∫tan-1xx2 dx

Q: Evaluate the integral dx fax √2x-x1

A: Given: I=∫dx2x-x2   As we know the common integral; ∫11-x2dx=sin-1x+C

Q: Findetact valves Sin (e) = casCe)=. tanló)= CSCG)= Secco)= CotHe= 29 %3D 45

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Q: 5 Find xy dydx.

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Q: 8 Evelluate the double integral: + 3x) dydx

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Q: 7 show that: sech x dx = Sin (tankx). √1-u² multiply note: * & Sim'u = dx * you Cam سال dx sechx by…

A: Differentiate rhs

Q: Find the area.   y = 3 cos(4x),    y = 3 − 3 cos(4x),    0 ≤ x ≤ π/4

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Q: Evaluate: 1-11-24 = - dx 1 + x 0 By the trapezoidal rule with h = 0.5, 0.25 and 0.125 respectively.

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Q: r(10,3,3) dx (1,1,1) dy dz. - -

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Q: 2. [2x*tan y – ysin (2) – 1] dx + [xsin (2) + x*sec*y] dy =

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Q: d (csc x) = -csc x cot x. dx Prove that

A: Given,          ddxcscx=-cscxcotx

Q: Evaluate the integral. 1//3 3earctan(y) dy -V3 1 + y2

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Q: 2. Evaluade integral.. iterated double ze^ dx dy

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Q: dx - Evaluate V4x' +9

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Q: Sx"dx

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Q: Find dy. y = 5 In (5+ x°) dy =

A: If you like the solution then please give it a thumbs up.... The Answer is:           Result :…

Q: Show that the antiderivatives are equivalent.                                                     ∫…

A: To show that ∫3x21-x6dx = arc sin x3+C Substitute  u = x3. Then                  dudx=3x2du  =3x2dx…

Q: Evaluate 2Sin x´dx, by (i) mid-point rule, and (ii) Simpson's 1/3 rule.

A: i)Given:        ∫242sinx2dx   Mid-point rule:       ∫abf(x)dx=∆xfx0+x12+fx1+x22+fx2+x32+...…

Q: TO Show Jv - x-V そーレ V+ dx

A: Jvx = ∑m=0∞ -1mx22m+vm! m+v!

Q: Vx - 2 dx

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Q: 2. Apply appropriate trigonometric substitution to compute the integral 1-4x +2 dx /2x

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Q: sec? (4x)(3 – tan(4x))³dx

A: Given integral is,  ∫sec24x3-tan4x3dx. Substitute t=3-tan4x.…

Q: vE zy sin{z° + y)dzdy Find

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Q: Evaluate double integral ∫20∫1y/2cos(x2)dx dy

A: According to the given information it is required to evaluate the integral

Q: sE de = 6 - +- + arctan( )) +c %3D 1+ Vx

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Q: 3. Stanh3x dx = Jsinh3x, Ssinh3x_ dx %3D u = cosh3x du = %3D

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Q: Iterated integrais. 3. ( ye* dx dy Jo Jo

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Q: 3. Approximate the integral -dx using 4 trapezoids.

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Q: 1 point) Calculate the integral: 1 dx = (x + 8)(x + 4)

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Q: If y=tanx , prove that y```=2(y^2+1)(3y^2+1)

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Q: dx Evaluate

A: ∫0919+x2dx

Q: Evaluate by Integration by parts (IBP): 4. x° cos 2x dx

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Q: 2T COs x dx

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Q: 2olue dy %3D

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Q: The area under one arch of y=sin(0.5x) is O 0.2 O 0.5 O 0.8 None of them

A: Given: y=sin(0.5x)

Q: find the area of the shaded 4) reglon y = 2 2- y = 2 sin(Tx) 3 х -1- -2+ -3+

A: From the given graph the shaded part is on the interval x=0.5 to 2.5.  So find the area between the…

Q: Evaluate the double integral. 5 25- х2 dy dx Jo

A: First, integrate with respect to y

Q: Use integration by substitution or integration by parts as appropriate to evaluate Axsin (Bx) dx…

A: We are given A=3 and B=3 now putting the values in integral it becomes ∫0π23xsin(3x) dx first we…

Q: O Use integration by part ax Cosbx dx

A:

Q: (a) In(4x) dx

A: Integration by parts:  ∫f(x)g′(x)dx = f(x)g(x) −∫f′(x)g(x)dx