हिंदी

Ed∫0π2 cosxesinx dx is equal to ______.

Advertisements
Advertisements

प्रश्न

`int_0^(pi/2)  cos x "e"^(sinx)  "d"x` is equal to ______.

रिक्त स्थान भरें
Advertisements

उत्तर

`int_0^(pi/2)  cos x "e"^(sinx)  "d"x` is equal to e – 1.

Explanation:

Let I = `int_0^(pi/2)  cos x "e"^(sinx)  "d"x` 

Put sin x = t

⇒ cos x "d"x` = dt

∴ I = `int_0^1 "e"^"t"  "dt"`

= `["e"^"t"]_0^1`

= `"e"^1 - "e"^0`

= e – 1

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Integrals - Exercise [पृष्ठ १६९]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 7 Integrals
Exercise | Q 59 | पृष्ठ १६९

संबंधित प्रश्न

 
 

Evaluate `int_(-2)^2x^2/(1+5^x)dx`

 
 

Evaluate: `int_(-a)^asqrt((a-x)/(a+x)) dx`


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

`int_0^(pi/2) cos^2 x dx`


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

`int_0^a  sqrtx/(sqrtx + sqrt(a-x))   dx`


`int_(-pi/2)^(pi/2) (x^3 + x cos x + tan^5 x + 1) dx ` is ______.


Prove that `int _a^b f(x) dx = int_a^b f (a + b -x ) dx`  and hence evaluate   `int_(pi/6)^(pi/3) (dx)/(1 + sqrt(tan x))` .   


Find `dy/dx, if y = cos^-1 ( sin 5x)`


`int_2^7 sqrt(x)/(sqrt(x) + sqrt(9 - x))  dx` = ______.


Choose the correct alternative:

`int_(-9)^9 x^3/(4 - x^2)  "d"x` =


Evaluate `int_0^1 x(1 - x)^5  "d"x`


`int (cos x + x sin x)/(x(x + cos x))`dx = ?


`int_0^(pi/4) (sec^2 x)/((1 + tan x)(2 + tan x))`dx = ?


`int_0^1 ((x^2 - 2)/(x^2 + 1))`dx = ?


`int_0^4 1/(1 + sqrtx)`dx = ______.


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


`int_0^(pi/2) sqrt(cos theta) * sin^2 theta "d" theta` = ______.


`int_0^{pi/2} cos^2x  dx` = ______ 


`int_-1^1x^2/(1+x^2)  dx=` ______.


`int_(-pi/4)^(pi/4) 1/(1 - sinx) "d"x` = ______.


`int_(-1)^1 (x^3 + |x| + 1)/(x^2 + 2|x| + 1) "d"x` is equal to ______.


Evaluate the following:

`int_0^(pi/2)  "dx"/(("a"^2 cos^2x + "b"^2 sin^2 x)^2` (Hint: Divide Numerator and Denominator by cos4x)


Evaluate: `int_((-π)/2)^(π/2) (sin|x| + cos|x|)dx`


The value of `int_((-1)/sqrt(2))^(1/sqrt(2)) (((x + 1)/(x - 1))^2 + ((x - 1)/(x + 1))^2 - 2)^(1/2)`dx is ______.


If `lim_("n"→∞)(int_(1/("n"+1))^(1/"n") tan^-1("n"x)"d"x)/(int_(1/("n"+1))^(1/"n") sin^-1("n"x)"d"x) = "p"/"q"`, (where p and q are coprime), then (p + q) is ______.


`int_(π/3)^(π/2) x sin(π[x] - x)dx` is equal to ______.


`int_0^(π/2)((root(n)(secx))/(root(n)(secx + root(n)("cosec"  x))))dx` is equal to ______.


`int_-1^1 |x - 2|/(x - 2) dx`, x ≠ 2 is equal to ______.


Evaluate: `int_0^(π/4) log(1 + tanx)dx`.


Solve the following.

`int_1^3 x^2 logx  dx`


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


Evaluate: `int_-1^1 x^17.cos^4x  dx`


Evaluate the following integral:

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


Evaluate the following integral:

`int_0^1 x(1-x)^5 dx`


Evaluate the following integral:

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


Evaluate the following definite intergral:

`int_1^3logx  dx`


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


Which property is useful for piecewise functions or modulus (absolute value) functions?


Under which condition does \[P_6\] give \[\int_{0}^{2a} f(x)\,dx=2\int_{0}^{a} f(x)\,dx\]?


Why can \[\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sin^{2}x\,dx\] be written as \[2\int_{0}^{\frac{\pi}{4}}\sin^{2}x\,dx\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×