Jx sinx cosx dx

Q: S. | sech (Insinx) cotxdx=2tan - (sinx) + C

A:

Q: fex cos (2x) dx

A:

Q: Sx tan=1x dx

A:

Q: find dy/dx, y=tanx-secx

A:

Q: Csx 7. J dx 5+sinx

A: We can  solve by using Antidifferentiation property

Q: Evaluate: J 1+sinx -dx

A: To evaluate the given integral.

Q: sinx dx TI cosx t ecoSX 2

A:

Q: y=(sinx)cosx

A:

Q: dy Solve dx ex +

A:

Q: S(sinx – cosx)²dx COoSx

A: Given, ∫sinx-cosx2dx  We have to find the integral.

Q: S(Sinx +CoSX) 人×

A: Solution: The objective is to find the value of the given integral

Q: Q6 Evaluate: sinx dx 1+ sinx

A: Given, ∫sinx1+sinxdx  We have to evaluate the integral.

Q: y' = Jy + sinx – cosx

A:

Q: find dy/dx, y= ln(cosecx-cotx)

A: given dydx=lncosecx-cotx

Q: S sec²x(2cosx + sinx)²dx

A:

Q: Find dy/dx:

A: We have to solve the dy/dx using fundamental theorem of calculus.

Q: Seax sin bx dx

A:

Q: Find Jxe* dx use parts

A:

Q: x' sinx dx

A:

Q: d sin x dx

A: We will use the quotient rule.  Here, u=sinx and v=x

Q: Socosx sinx dx X.

A: Given definite integral is

Q: sinx + cosx dx cosx

A: We have to evaluate the integral                                                    ∫sinx+cosxcosxdx

Q: Find dy /dx fors catx y-(tanx) +Ccata) (cotA) tanx

A: We use Logarithmic properties and differentiation

Q: find dy dx

A: Given equation is x2+xy=1 We have to find dydx Differentiate  x2+xy=1 both sides by x…

Q: 9. Sx* Sinx dx

A: To Evaluate: ∫x3sinxdx Concept: Integration By Parts:∫udv=uv-∫vdu

Q: tanx sec?x dx

A: Let,I=∫tanxsec2xdxLet,tanx=tsec2xdx=dttherefore,I=∫tdtI=t22+CNow,substituting value of…

Q: Find dy dx

A: The given equation is y3+y2-5y-x2=-4. Differentiate both sides with respect to x as follows.

Q: Sinx dx

A: Here we have to calculate the integral in the given limit: ∫−π6π6x6sinx1+x10dxwhere   fx=x6sinx1+x10

Q: s sinzx cosx dx

A:

Q: 5x sinx dx = %3D

A:

Q: Find dy / dx

A: (As per bartleby guidelines, for multiple questions asked, only the first question will be solved.…

Q: f sech³x tanhx dx

A:

Q: Sx.Sinx Cosx.dx

A:

Q: sinx dx

A:

Q: cosx dx by u

A: Given: ∑n=1∞2nn! ∫cos8x dx

Q: y = cosx sinx

A: Given query is to the derivative of the expression.

Q: Jx' sin x dx

A: we need to evaluate, ∫x2 sin x dx this integral can be done by integration by parts. ∫u·v dx = u·∫v…

Q: Sx sin2x dx

A: I am integrating the above term by integration by parts method.

Q: Find dy/dx

A: (1) Given function is: y = sqrt {e^(2x) + cot x} We have to find the derivative dy/dx.…

Q: dx 1+cosx

A: Here we have to evaluate the given integration.

Q: find dy/dx,y= cotx-tanx

A: Given function is Derivative of cotx is -(cscx)^2 and derivative of tanx is (secx)^2 Therefore,

Q: 0 sinx + 2cosx dx

A:

Q: y(cscx-cotx) =x- sinx + c

A:

Q: Cos (Jx) dx

A:

Q: Calculate x sin(y)dxdy

A: On solving this we will get

Q: गयहिनाड ते coSx+cosy Cotx dx CSC X

A:

Q: y = Secx ( sinx+ Cosx)

A: Given, y=secxsinx+cosx And we have to find the dydx.

Q: Se X ết Sinrd Secose da dx

A: We will use integration by parts to solve the integral .  ∫uv dx=u∫vdx -∫dudx∫v dxdx

Q: d „sin x dx

A:

Q: eMx cos nx dx

A:

Rate of Change

The relation between two quantities which displays how much greater one quantity is than another is called ratio.

Slope

The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as: