मराठी

By using the properties of the definite integral, evaluate the integral: ∫π2π2sin7xdx

Advertisements
Advertisements

प्रश्न

By using the properties of the definite integral, evaluate the integral:

`int_(pi/2)^(pi/2) sin^7 x dx`

बेरीज
Advertisements

उत्तर

Let f (x) = sin7 x.

sin x is an odd function

i.e. if h (x) = sin x

⇒ h (-x) = sin (-x)

= - sin (x) = -h (x)

⇒ odd power of sin x is odd

⇒ f (x) is an odd function of x.

⇒ `int_(-pi/2)^(pi/2) sin^7 x  dx = 0`        .... [∵ If f (x) is odd ⇒`int_-a^a` f (x) dx = 0]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Integrals - Exercise 7.11 [पृष्ठ ३४७]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
पाठ 7 Integrals
Exercise 7.11 | Q 13 | पृष्ठ ३४७

संबंधित प्रश्‍न

 
 

Evaluate : `intlogx/(1+logx)^2dx`

 
 

By using the properties of the definite integral, evaluate the integral:

`int_(-5)^5 | x + 2| dx`


By using the properties of the definite integral, evaluate the integral:

`int_((-pi)/2)^(pi/2) sin^2 x  dx`


Evaluate : \[\int(3x - 2) \sqrt{x^2 + x + 1}dx\] .


Evaluate  : `int "x"^2/("x"^4 + 5"x"^2 + 6) "dx"`


Evaluate = `int (tan x)/(sec x + tan x)` . dx


Prove that `int_0^"a" "f" ("x") "dx" = int_0^"a" "f" ("a" - "x") "d x",` hence evaluate `int_0^pi ("x" sin "x")/(1 + cos^2 "x") "dx"`


`int_"a"^"b" "f"(x)  "d"x` = ______


Choose the correct alternative:

`int_(-9)^9 x^3/(4 - x^2)  "d"x` =


Evaluate `int_1^3 x^2*log x  "d"x`


`int_0^{pi/2} xsinx dx` = ______


If `int_0^"a" sqrt("a - x"/x) "dx" = "K"/2`, then K = ______.


`int_0^{pi/4} (sin2x)/(sin^4x + cos^4x)dx` = ____________


`int_0^{pi/4} (sin2x)/(sin^4x + cos^4x)dx` = ____________


`int_0^{1/sqrt2} (sin^-1x)/(1 - x^2)^{3/2} dx` = ______ 


`int_0^1 log(1/x - 1) "dx"` = ______.


`int_(pi/4)^(pi/2) sqrt(1-sin 2x)  dx =` ______.


Show that `int_0^(pi/2) (sin^2x)/(sinx + cosx) = 1/sqrt(2) log (sqrt(2) + 1)`


`int_("a" + "c")^("b" + "c") "f"(x) "d"x` is equal to ______.


`int_0^(2"a") "f"(x) "d"x = 2int_0^"a" "f"(x) "d"x`, if f(2a – x) = ______.


`int_0^(pi/2) sqrt(1 - sin2x)  "d"x` is equal to ______.


If `int_0^"a" 1/(1 + 4x^2) "d"x = pi/8`, then a = ______.


`int_0^(2"a") "f"("x") "dx" = int_0^"a" "f"("x") "dx" + int_0^"a" "f"("k" - "x") "dx"`, then the value of k is:


Evaluate:

`int_2^8 (sqrt(10 - "x"))/(sqrt"x" + sqrt(10 - "x")) "dx"`


The value of the integral `int_0^sqrt(2)([sqrt(2 - x^2)] + 2x)dx` (where [.] denotes greatest integer function) is ______.


`int_0^(pi/4) (sec^2x)/((1 + tanx)(2 + tanx))dx` equals ______.


Evaluate: `int_1^3 sqrt(x + 5)/(sqrt(x + 5) + sqrt(9 - x))dx`


Assertion (A): `int_2^8 sqrt(10 - x)/(sqrt(x) + sqrt(10 - x))dx` = 3.

Reason (R): `int_a^b f(x) dx = int_a^b f(a + b - x) dx`.


Evaluate `int_0^3root3(x+4)/(root3(x+4)+root3(7-x))  dx`


Evaluate the following definite integral:

`int_1^3 log x  dx`


Evaluate the following integral:

`int_-9^9 x^3/(4 - x^2) dx`


Solve the following.

`int_0^1e^(x^2)x^3 dx`


Evaluate the following integral:

`int_-9^9x^3/(4-x^2)dx`


Solve the following.

`int_0^1e^(x^2)x^3dx`


Solve the following.

`int_0^1e^(x^2)x^3dx`


\[\int_{-2}^{2}\left|x^{2}-x-2\right|\mathrm{d}x=\]


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


Which expression is equal to \[\frac{1}{1+\sqrt{\tan x}}\] in the evaluation of \[\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{dx}{1+\sqrt{\tan x}}\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×