मराठी

Evaluate the definite integral: ∫0π4sinx+ cosx9+16sin2xdx - Mathematics

Advertisements
Advertisements

प्रश्न

Evaluate the definite integral:

`int_0^(pi/4) (sin x +  cos x)/(9+16sin 2x) dx`

बेरीज
Advertisements

उत्तर

Let `I = int_0^(pi/4) (sin x + cos x)/(9 + 16 sin 2x)`dx

Put sin x - cos x = t 

(cos x + sin x)dx = dt

and 1 - 2 sin x cos x = t2

⇒ sin 2x = 1 - t2

When x = `pi/4`, t = sin `pi/4 - cos  pi/4`

`= 1/sqrt2 - 1/sqrt2 = 0`

When x = 0, t = sin 0 - cos 0 = - 1

`therefore int_0^(pi/4) (sin x + cos x)/(9 + 16 sin 2x)`dx

`= int_(- 1)^0 dt/(9 + 16 (1 - t^2))`

`= int_(- 1)^0 dt/(25 - 16 t^2)`

`= 1/16 int_(- 1)^0 dt/((5/4)^2 - t^2)`

`= 1/16 * 1/(2 * 5/4) [log |(5/4 + t)/(5/4 - t)|]_(-1)^0`

`= 1/40 [log 1 - (log 1 - log 9)]`

`= 1/40 log 9`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Integrals - Exercise 7.12 [पृष्ठ ३५३]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
पाठ 7 Integrals
Exercise 7.12 | Q 30 | पृष्ठ ३५३

व्हिडिओ ट्यूटोरियलVIEW ALL [3]

संबंधित प्रश्‍न

Evaluate `int_1^3(e^(2-3x)+x^2+1)dx`  as a limit of sum.


Evaluate the following definite integrals as limit of sums.

`int_0^5 (x+1) dx`


Evaluate the following definite integrals as limit of sums.

`int_1^4 (x^2 - x) dx`


Evaluate the following definite integrals as limit of sums.

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


Evaluate the definite integral:

`int_(pi/2)^pi e^x ((1-sinx)/(1-cos x)) dx`


Evaluate the definite integral:

`int_0^(pi/4) (sinx cos x)/(cos^4 x + sin^4 x)`dx


Evaluate the definite integral:

`int_0^(pi/2) (cos^2 x dx)/(cos^2 x + 4 sin^2 x)`


Evaluate the definite integral:

`int_(pi/6)^(pi/3)  (sin x + cosx)/sqrt(sin 2x) dx`


Evaluate the definite integral:

`int_0^(pi/2) sin 2x tan^(-1) (sinx) dx`


Prove the following:

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


Prove the following:

`int_0^(pi/2) sin^3 xdx = 2/3`


Prove the following:

`int_0^(pi/4) 2 tan^3 xdx = 1 - log 2`


Evaluate  `int_0^1 e^(2-3x) dx` as a limit of a sum.


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`


` ∫  log x / x  dx `
 
 
 

\[\int\frac{\sin^3 x}{\sqrt{\cos x}} dx\]

\[\int\frac{1}{\sqrt{\tan^{- 1} x} . \left( 1 + x^2 \right)} dx\]

\[\int\frac{1}{x} \left( \log x \right)^2 dx\]


\[\int\frac{\sin x}{\left( 1 + \cos x \right)^2} dx\]

 


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

\[\int\log x\frac{\text{sin} \left\{ 1 + \left( \log x \right)^2 \right\}}{x} dx\]

\[\int \sec^4    \text{ x   tan x dx} \]

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

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

 


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


Evaluate:

`int (sin"x"+cos"x")/(sqrt(9+16sin2"x")) "dx"`


Evaluate the following:

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


Evaluate the following:

`int_0^(1/2) ("d"x)/((1 + x^2)sqrt(1 - x^2))`  (Hint: Let x = sin θ)


Evaluate the following:

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


The value of `lim_(x -> 0) [(d/(dx) int_0^(x^2) sec^2 xdx),(d/(dx) (x sin x))]` is equal to


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×