English
Tamil Nadu Board of Secondary EducationHSC Science Class 12

Evaluate the following integrals using properties of integration: tdttdt∫0sin2xsin-1t dt+∫0cos2xcos-1t dt - Mathematics

Advertisements
Advertisements

Question

Evaluate the following integrals using properties of integration:

`int_0^(sin^2x) sin^-1 sqrt("t")  "dt" + int_0^(cos^2x) cos^-1 sqrt("t")  "dt"`

Sum

Solution

I1 = `int_0^(sin^2x) sin^-1 sqrt("t")  "dt"`

Put `si^-1 sqrt("'t") = theta`

`sqrt("t")` = sin θ

t sin2x
θ 0 x

`1/(2sqrt("t")) "dt"` = cos θ dθ

dt = `2sqrt("t") cos theta  "d"theta`

= 2 sin θ cos θ dθ

dt = sin 2θ dθ

I1 = `int_0^x theta  sin2theta  "d"theta`

= `[ (- thetacos2theta)/2 + (sin2theta)/4]_0^x`

= `(-x cos 2x)/2 + (sin2x)/4`  ......(1)

I1 =  `int_0^(cos^2x)  cos^-1 sqrt("t") "dt"`

Put `cos^-1 sqrt("t")` = θ

`sqrt("t"")` = cos θ

`1/(2sqrt("t"))  "dt"` = – sin θ dθ

dt`- 2sqrt("t")  sin theta  "d"theta`

= – 2cos θ sin θ dθ
dt = – sin 2θ dθ

I2 = `int_0^(cos^x) cos^-1  sqrt("t")  "dt"`

= `int_(pi/2)^x - theta sin 2theta  "d"theta`

t cos2x
θ `pi/2` x

= `- [(- theta cos 2theta)/2+ (sin theta)/4]_(pi/2)^x`

= `[(theta cos2theta)/2 - (sin 2theta)/4]_(p/2)^x`

= `(x cos 2x)/2 - (sin 2x)/4 + pi/4`   ........(2)

I = I1 + I2

= `(- x cos 2x)/2 + (sin 2x)/4 + (x cos2x)/2 - (sin 2x)/4 + pi/4`

I = `pi/4`

shaalaa.com
Fundamental Theorems of Integral Calculus and Their Applications
  Is there an error in this question or solution?
Chapter 9: Applications of Integration - Exercise 9.3 [Page 113]

APPEARS IN

Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 12 TN Board
Chapter 9 Applications of Integration
Exercise 9.3 | Q 2. (vii) | Page 113

RELATED QUESTIONS

Evaluate the following definite integrals:

`int_3^4 (d"x)/(x^2 - 4)`


Evaluate the following definite integrals:

`int_(-1)^1 ("d"x)/(x^2 + 2x + 5)`


Evaluate the following definite integrals:

`int_0^1 sqrt((1 - x)/(1 + x)) "d"x`


Evaluate the following definite integrals:

`int_0^(pi/2) "e"^x((1 + sin x)/(1 + cos x))"d"x`


Evaluate the following definite integrals:

`int_0^1 (1 - x^2)/(1 + x^2)^2  "d"x`


Evaluate the following integrals using properties of integration:

`int_(-5)^5 x cos(("e"^x - 1)/("e"^x + 1))  "d"x`


Evaluate the following integrals using properties of integration:

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


Evaluate the following integrals using properties of integration:

`int_(- pi/4)^(pi/4) sin^2x  "d"x`


Evaluate the following integrals using properties of integration:

`int_0^(2pi) x log((3 + cosx)/(3 - cosx)) "d"x`


Evaluate the following integrals using properties of integration:

`int_0^pi sin^4 x cos^3 x  "d"x`


Evaluate the following integrals using properties of integration:

`int_0^1 |5x - 3|  "d"x`


Evaluate the following integrals using properties of integration:

`int_0^1 (log(1 + x))/(1 + x^2)  "d"x`


Evaluate the following integrals using properties of integration:

`int_(pi/8)^((3pi)/8) 1/(1 + sqrt(tan x))  "d"x`


Evaluate the following integrals using properties of integration:

`int_0^pi x[sin^2(sin x) cos^2 (cos x)] "d"x`


Choose the correct alternative:

For any value of n ∈ Z, `int_0^pi "e"^(cos^2x) cos^3[(2n+ 1)x]  "d"x` is


Choose the correct alternative:

The value of `int_(-4)^4 [tan^-1  ((x^2)/(x^4 + 1)) + tan^-1 ((x^4 + 1)/x^2)] "d"x` is


Choose the correct alternative:

The value of `int_(- pi/4)^(pi/4) ((2x^7 - 3x^5 + 7x^3 - x + 1)/(cos^2x)) "d"x` is


Choose the correct alternative:

The value of `int_0^pi ("d"x)/(1 + 5^(cosx))` is


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×