English

Solve: dy/dx = cos(x + y)

Advertisements
Advertisements

Question

Solve:

dy/dx = cos(x + y)

Sum
Advertisements

Solution

Given,

`dy/dx= cos (x + y)` …(i)

Put `x + y = v`        …(ii)

`∴ y = v – x`

`∴ dy/dx=(dv)/dx-1`  …(iii)

Substituting (ii) and (iii) in (i), we get

`(dv)/dx-1=cosv`

`therefore (dv)/dx=1+cosv`

`therefore (dv)/dx=2cos^2(v/2)`

`therefore 1/cos^2(v/2)dv=2dx`

`therefore sec^2(v/2)dv=2dx`

Integrating on both sides, we get

`int sec^2(v/2)dv=2intdx`

`therefore 2tan(v/2)=2x+c'`

`therefore tan(v/2)=x+(c')/2`

`therefore tan((x+y)/2)=x+c`, where `c=(c')/2`

shaalaa.com
  Is there an error in this question or solution?
2016-2017 (March)

APPEARS IN

RELATED QUESTIONS

Evaluate : `int (sinx)/sqrt(36-cos^2x)dx`


Integrate the functions:

`xsqrt(x + 2)`


Integrate the functions:

`e^(2x+3)`


Integrate the functions:

`x/(e^(x^2))`


\[\int\sqrt{x^2 + x + 1} \text{ dx}\]

Write a value of

\[\int \tan^3 x \sec^2 x \text{ dx }\].

 


Write a value of\[\int\frac{\sin 2x}{a^2 \sin^2 x + b^2 \cos^2 x} \text{ dx }\]


\[If \int e^x \left( \tan x + 1 \right)\text{ sec  x  dx } = e^x f\left( x \right) + C, \text{ then  write  the value  of  f}\left( x \right) .\]

 

 


 Prove that: `int "dx"/(sqrt("x"^2 +"a"^2)) = log  |"x" +sqrt("x"^2 +"a"^2) | + "c"`


Integrate the following w.r.t. x : x3 + x2 – x + 1


Integrate the following functions w.r.t. x : `(logx)^n/x`


Integrate the following functions w.r.t. x : `((sin^-1 x)^(3/2))/(sqrt(1 - x^2)`


Integrate the following functions w.r.t. x : sin5x.cos8x


Integrate the following functions w.r.t. x : `int (1)/(3 + 2 sin2x + 4cos 2x).dx`


Evaluate the following integrals :  `int (3x + 4)/sqrt(2x^2 + 2x + 1).dx`


Choose the correct options from the given alternatives : 

`int dx/(cosxsqrt(sin^2x - cos^2x))*dx` =


Choose the correct options from the given alternatives :

`int (cos2x - 1)/(cos2x + 1)*dx` =


If f'(x) = x2 + 5 and f(0) = −1, then find the value of f(x).


Evaluate the following.

`int 1/(4"x"^2 - 1)` dx


Evaluate the following.

`int 1/(sqrt(3"x"^2 - 5))` dx


Evaluate: `int "e"^sqrt"x"` dx


`int 1/(cos x - sin x)` dx = _______________


`int (2(cos^2 x - sin^2 x))/(cos^2 x + sin^2 x)` dx = ______________


If `int 1/(x + x^5)` dx = f(x) + c, then `int x^4/(x + x^5)`dx = ______


`int(log(logx))/x  "d"x`


`int dx/(1 + e^-x)` = ______


`int ((x + 1)(x + log x))^4/(3x) "dx" =`______.


`int ("e"^x(x + 1))/(sin^2(x"e"^x)) "d"x` = ______.


The value of `intsinx/(sinx - cosx)dx` equals ______.


Evaluate `int_-a^a f(x) dx`, where f(x) = `9^x/(1 + 9^x)`.


Evaluate the following.

`int(20 - 12"e"^"x")/(3"e"^"x" - 4) "dx"`


Evaluate:

`int sin^3x cos^3x  dx`


Evaluate:

`int(5x^2-6x+3)/(2x-3)dx`


Evaluate the following.

`int "x"^3/sqrt(1 + "x"^4)` dx


Evaluate the following.

`int1/(x^2+4x-5)dx`


What is integration by substitution?


Which standard substitution is used for \[\sqrt{\mathrm{a}^2-x^2}\], \[\frac{1}{\sqrt{\mathrm{a}^2-x^2}}\], or \[\mathrm{a}^2-x^2\]?


For \[\int\frac{\sin x}{\sin(x+a)}\,dx\], which substitution gives \[dx=dt\]?


How should a substitution be chosen?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×