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

Integrate the following functions w.r.t. x : tan 3x tan 2x tan x

Advertisements
Advertisements

प्रश्न

Integrate the following functions w.r.t. x :  tan 3x tan 2x tan x

बेरीज
Advertisements

उत्तर

Let I = `int tan 3x tan 2x tanx dx`
Consider tan 3x = tan (2x + x)

= `(tan2x + tanx)/(1 - tan2x tanx)`

∴ tan3x (1 – tan 2x tan x) = tan 2x + tan x
∴ tan 3x –  tan 3x tan 2x tan x = tan 2x + tan x
∴ tan 3x –  tan 2x –  tan x = tan 3x tan 2x tan x
I = `int (tan3x - tan 2x - tanx)dx`

= `int tan3xdx - int tan2x dx - int tanx dx`

= `(1)/(3)log|sec3x| - (1)/(2)log|sec2x| - log|secx| + c`.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Indefinite Integration - Exercise 3.2 (A) [पृष्ठ ११०]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 3 Indefinite Integration
Exercise 3.2 (A) | Q 2.13 | पृष्ठ ११०

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

Integrate the functions:

`e^(tan^(-1)x)/(1+x^2)`


Integrate the functions:

`sqrt(sin 2x) cos 2x`


Evaluate `int (x-1)/(sqrt(x^2 - x)) dx`


Evaluate: `int (sec x)/(1 + cosec x) dx`


\[\int\sqrt{9 - x^2}\text{ dx}\]

Write a value of

\[\int e^x \left( \sin x + \cos x \right) \text{ dx}\]

 


Write a value of

\[\int\frac{a^x}{3 + a^x} \text{ dx}\]

Write a value of\[\int\sqrt{4 - x^2} \text{ dx }\]


Write a value of\[\int\sqrt{9 + x^2} \text{ dx }\].


\[\int x \sin^3 x\ dx\]

`int "dx"/(9"x"^2 + 1)= ______. `


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(x - 2)/sqrt(x + 5).dx`


Evaluate the following integrals : `int cos^2x.dx`


If `f'(x) = x - (3)/x^3, f(1) = (11)/(2)`, find f(x)


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

`x^5sqrt(a^2 + x^2)`


Integrate the following functions w.r.t. x : `sin(x - a)/cos(x  + b)`


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

cos8xcotx


Evaluate the following : `int (1)/sqrt(2x^2 - 5).dx`


Evaluate the following : `int sqrt((2 + x)/(2 - x)).dx`


Evaluate the following : `int (1)/sqrt(3x^2 + 5x + 7).dx`


Evaluate the following : `int (1)/sqrt(8 - 3x + 2x^2).dx`


Integrate the following functions w.r.t. x : `int (1)/(4 - 5cosx).dx`


Evaluate the following.

`int 1/("x" log "x")`dx


Fill in the Blank.

`int 1/"x"^3 [log "x"^"x"]^2 "dx" = "P" (log "x")^3` + c, then P = _______


Evaluate: `int sqrt("x"^2 + 2"x" + 5)` dx


`int x^2/sqrt(1 - x^6)` dx = ________________


`int sqrt(x^2 + 2x + 5)` dx = ______________


`int sqrt(x)  sec(x)^(3/2) tan(x)^(3/2)"d"x`


`int cot^2x  "d"x`


`int "e"^(sin^-1 x) ((x + sqrt(1 - x^2))/(sqrt1 - x^2)) "dx" = ?`


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


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


The integral `int ((1 - 1/sqrt(3))(cosx - sinx))/((1 + 2/sqrt(3) sin2x))dx` is equal to ______.


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


Evaluate the following.

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


Evaluate the following.

`intx sqrt(1 +x^2)  dx`


Evaluate the following:

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


Evaluate the following.

`int1/(x^2 + 4x - 5)  dx`


Evaluate the following.

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


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


Which substitution is appropriate for \[\sqrt{\frac{x}{a-x}}\], \[\sqrt{\frac{a-x}{x}}\], \[\sqrt{x(a-x)}\], or \[\frac{1}{\sqrt{x(a-x)}}\]?


For \[\sqrt{\frac{x-\alpha}{\beta-x}}\] or \[\sqrt{(x-\alpha)(\beta-x)}\], where \[\beta>\alpha\], which substitution is used?


After putting \[t=\cos x\], which integral is obtained from \[\int\sin^2x\cos^2x(\sin x)\,dx\]?


How should a substitution be chosen?


When applying substitution, what must always be rewritten in terms of the new variable?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×