हिंदी

Evaluate the following: md∫0π2tanx1+m2tan2xdx - Mathematics

Advertisements
Advertisements

प्रश्न

Evaluate the following:

`int_0^(pi/2) (tan x)/(1 + "m"^2 tan^2x) "d"x`

योग
Advertisements

उत्तर

Let I = `int_0^(pi/2) (tan x)/(1 + "m"^2 tan^2x) "d"x`

= `int_0^(pi/2) (sinx/cosx)/(1 + "m"^2 (sin^2x)/(cos^2x)) "d"x`

= `int_0^(pi/2) (sinx/cosx)/((cos^2x + "m"^2 sin^2x)/cos^2x) "d"x`

= `int_0^(pi/2) (sin x cos x)/(cos^2x + "m"^2 sin^2x) "d"x`

= `int_0^(pi/2) (sinx cosx)/(1 - sin^2x + "m"^2 sin^2x) "d"x`

= `int_0^(pi/2) (sinx cosx)/(1 - sin^2x (1 - "m"^2)) "d"x`

Put sin2x = t

2 sin x cos x dx = dt

sin x cos x dx = `"dt"//2`

Changing the limits we get,

When x = 0

∴ t = sin20 = 0

When x = `pi/2`

∴ t = `sin^2  pi/2` = 1

∴ I = `1/2 int_0^1  "dt"/(1 - (1 - "m"^2)"t")`

I = `1/2 int_0^1 "dt"/(1 + ("m"^2 - 1)"t")`

= `1/2 [(log [1 + "m"^2 - 1)"t")/("m"^2 - 1)]_0^1`

= `1/(2("m"^2 - 1)) [log(1 + "m"^2 - 1) - log(1)]`

= `(log|"m"^2|)/(2("m"^2 - 1))`

Hence, I = `(log|"m"^2|)/(2("m"^2 - 1)) = (log|"m"|)/("m"^2 - 1)`.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Integrals - Exercise [पृष्ठ १६५]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 7 Integrals
Exercise | Q 30 | पृष्ठ १६५

वीडियो ट्यूटोरियलVIEW ALL [3]

संबंधित प्रश्न

Evaluate `int_(-1)^2(e^3x+7x-5)dx` as a limit of sums


Evaluate the following definite integrals as limit of sums. 

`int_2^3 x^2 dx`


Evaluate the following definite integrals as limit of sums.

`int_0^4 (x + e^(2x)) dx`


Prove the following:

`int_0^1 xe^x dx = 1`


Prove the following:

`int_(-1)^1 x^17 cos^4 xdx = 0`


Prove the following:

`int_0^1sin^(-1) xdx = pi/2 - 1`


`int dx/(e^x + e^(-x))` is equal to ______.


`int (cos 2x)/(sin x + cos x)^2dx` is equal to ______.


Choose the correct answers The value of `int_0^1 tan^(-1)  (2x -1)/(1+x - x^2)` dx is 

(A) 1

(B) 0

(C) –1

(D) `pi/4`


if `int_0^k 1/(2+ 8x^2) dx = pi/16` then the value of k is ________.

(A) `1/2`

(B) `1/3`

(C) `1/4`

(D) `1/5`


Evaluate : `int_1^3 (x^2 + 3x + e^x) dx` as the limit of the sum.


\[\int\frac{1 + \cos x}{\left( x + \sin x \right)^3} dx\]

\[\int\frac{\log x^2}{x} dx\]

\[\text{ ∫  cosec x  log}      \left( \text{cosec x} - \cot x \right) dx\]

\[\int x^3 \sin \left( x^4 + 1 \right) dx\]

\[\int\frac{1}{x\sqrt{x^4 - 1}} dx\]

\[\int\limits_0^1 \left( x e^x + \cos\frac{\pi x}{4} \right) dx\]

 


\[\int\limits_0^\pi \frac{\sin x}{\sin x + \cos x} dx\]

Evaluate the following integral:

\[\int\limits_{- 1}^1 \left| 2x + 1 \right| dx\]

\[\int\frac{\sqrt{\tan x}}{\sin x \cos x} dx\]


Evaluate `int_1^4 ( 1+ x +e^(2x)) dx` as limit of sums.


Evaluate `int_(-1)^2 (7x - 5)"d"x` as a limit of sums


Evaluate the following as limit of sum:

`int _0^2 (x^2 + 3) "d"x`


Evaluate the following:

`int_0^pi x sin x cos^2x "d"x`


Evaluate the following:

`int_(pi/3)^(pi/2) sqrt(1 + cosx)/(1 - cos x)^(5/2)  "d"x`


If f" = C, C ≠ 0, where C is a constant, then the value of `lim_(x -> 0) (f(x) - 2f (2x) + 3f (3x))/x^2` is


The limit of the function defined by `f(x) = {{:(|x|/x",", if x ≠ 0),(0",", "otherwisw"):}`


What is the derivative of `f(x) = |x|` at `x` = 0?


`lim_(x -> 0) (xroot(3)(z^2 - (z - x)^2))/(root(3)(8xz - 4x^2) + root(3)(8xz))^4` is equal to


The value of  `lim_(n→∞)1/n sum_(r = 0)^(2n-1) n^2/(n^2 + 4r^2)` is ______.


`lim_(n→∞){(1 + 1/n^2)^(2/n^2)(1 + 2^2/n^2)^(4/n^2)(1 + 3^2/n^2)^(6/n^2) ...(1 + n^2/n^2)^((2n)/n^2)}` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×