मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Integrate the following functions w.r.t. x : ∫1cosx-sinx.dx - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Integrate the following functions w.r.t. x : `int (1)/(cosx - sinx).dx`

बेरीज
Advertisements

उत्तर

Let I = `int (1)/(cosx - sinx).dx`

Dividing each term by `sqrt(1^2 + (-1)^2) = sqrt(2)`, we get

I = `(1)/sqrt(2) int (1)/(cosx. 1/sqrt(2) - sinx. 1/sqrt(2)).dx`

= `1/sqrt(2) int (1)/(cosx  . cos  pi/(4) - sin x. sin  pi/(4)).dx`

= `1/sqrt(2) int (1)/(cos(x + pi/4)).dx`

= `1/sqrt(2) int sec(x + pi/4).dx`

= `1/sqrt(2)log|sec(x + pi/4) + tan(x + pi/4)| + c`.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Indefinite Integration - Exercise 3.2 (B) [पृष्ठ १२३]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 3 Indefinite Integration
Exercise 3.2 (B) | Q 2.8 | पृष्ठ १२३

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

Integrate the functions:

`(2x)/(1 + x^2)`


Integrate the functions:

`(sin^(-1) x)/(sqrt(1-x^2))`


Integrate the functions:

`1/(cos^2 x(1-tan x)^2`


Integrate the functions:

cot x log sin x


Integrate the functions:

`1/(1 - tan x)`


Integrate the functions:

`sqrt(tanx)/(sinxcos x)`


Write a value of

\[\int\frac{1 + \cot x}{x + \log \sin x} \text{ dx }\]

Write a value of\[\int\frac{\sin x - \cos x}{\sqrt{1 + \sin 2x}} \text{ dx}\]


Integrate the following w.r.t. x:

`3 sec^2x - 4/x + 1/(xsqrt(x)) - 7`


Evaluate the following integrals : tan2x dx


Evaluate the following integrals : `int (sin2x)/(cosx)dx`


Evaluate the following integrals : `int sinx/(1 + sinx)dx`


Evaluate the following integrals : `int sqrt(1 + sin 2x) dx`


Evaluate the following integral: 

`int(4x + 3)/(2x + 1).dx`


Evaluate the following integrals : `int (3)/(sqrt(7x - 2) - sqrt(7x - 5)).dx`


Integrate the following functions w.r.t. x : `(x.sec^2(x^2))/sqrt(tan^3(x^2)`


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


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


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


Evaluate the following integrals:

`int (7x + 3)/sqrt(3 + 2x - x^2).dx`


Choose the correct options from the given alternatives :

`int f x^x (1 + log x)*dx`


Choose the correct options from the given alternatives :

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


Evaluate `int (3"x"^2 - 5)^2` dx


Evaluate the following.

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


Evaluate: `int "e"^sqrt"x"` dx


`int  ("e"^x(x - 1))/(x^2)  "d"x` = ______ 


`int "e"^x[((x + 3))/((x + 4)^2)] "d"x`


`int ("e"^(2x) + "e"^(-2x))/("e"^x)  "d"x`


`int 1/(xsin^2(logx))  "d"x`


`int cot^2x  "d"x`


`int sqrt(("e"^(3x) - "e"^(2x))/("e"^x + 1))  "d"x`


To find the value of `int ((1 + logx))/x` dx the proper substitution is ______


`int (1 + x)/(x + "e"^(-x))  "d"x`


If f(x) = 3x + 6, g(x) = 4x + k and fog (x) = gof (x) then k = ______.


General solution of `(x + y)^2 ("d"y)/("d"x) = "a"^2, "a" ≠ 0` is ______. (c is arbitrary constant)


If `int x^3"e"^(x^2) "d"x = "e"^(x^2)/2 "f"(x) + "c"`, then f(x) = ______.


`int 1/(a^2 - x^2) dx = 1/(2a) xx` ______.


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


If `int sinx/(sin^3x + cos^3x)dx = α log_e |1 + tan x| + β log_e |1 - tan x + tan^2x| + γ tan^-1 ((2tanx - 1)/sqrt(3)) + C`, when C is constant of integration, then the value of 18(α + β + γ2) is ______.


`int x/sqrt(1 - 2x^4) dx` = ______.

(where c is a constant of integration)


Evaluate `int_(logsqrt(2))^(logsqrt(3)) 1/((e^x + e^-x)(e^x - e^-x)) dx`.


Evaluate `int(1+ x + x^2/(2!)) dx`


`int dx/((x+2)(x^2 + 1))`    ...(given)

`1/(x^2 +1) dx = tan ^-1 + c`


`int (cos4x)/(sin2x + cos2x)dx` = ______.


Evaluate the following:

`int (1) / (x^2 + 4x - 5) dx`


Evaluate the following.

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


If f'(x) = 4x3 – 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x).


If f'(x) = 4x3 - 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x). 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×