हिंदी

Evaluate the following: dttt∫dt3t-2t2

Advertisements
Advertisements

प्रश्न

Evaluate the following:

`int "dt"/sqrt(3"t" - 2"t"^2)`

योग
Advertisements

उत्तर

Let I = `int "dt"/sqrt(3"t" - 2"t"^2)`

= `int "dt"/sqrt(-2("t"^2 - 3/2 "t"))`

= `1/sqrt(2) int "dt"/sqrt(-("t"^2 - 3/2 "t" + 9/16 - 9/16))` ....[Making perfect square]

= `1/sqrt(2) int "dt"/sqrt(-[("t" - 3/4)^2 - 9/16])`

= `1/sqrt(2) int "dt"/sqrt(9/16 - ("t" - 3/4)^2)`

= `1/sqrt(2) int "dt"/sqrt((3/4)^2 - ("t" - 3/4)^2)`

= `1/sqrt(2) * sin^-1  ("t" - 3/4)/(3/4) + "C"`

= `1/sqrt(2) sin^-1  (4"t" - 3)/3 + "C"`

Hence, I = `1/sqrt(2) sin^-1 ((4"t" - 3)/3) + "C"`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Integrals - Exercise [पृष्ठ १६४]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 7 Integrals
Exercise | Q 15 | पृष्ठ १६४

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Evaluate:  `int (1+logx)/(x(2+logx)(3+logx))dx`


 

Evaluate `∫_0^(3/2)|x cosπx|dx`

 

 

find `∫_2^4 x/(x^2 + 1)dx`

 

Evaluate :

`∫_0^π(4x sin x)/(1+cos^2 x) dx`


Evaluate the integral by using substitution.

`int_0^(pi/2) sqrt(sin phi) cos^5 phidphi`


Evaluate the integral by using substitution.

`int_0^2 dx/(x + 4 - x^2)`


Evaluate the integral by using substitution.

`int_1^2 (1/x- 1/(2x^2))e^(2x) dx`


If `f(x) = int_0^pi t sin  t  dt`, then f' (x) is ______.


Evaluate of the following integral:

(i)  \[\int x^4 dx\]

 


Evaluate of the following integral: 

\[\int\frac{1}{x^5}dx\]

Evaluate of the following integral: 

\[\int\frac{1}{x^{3/2}}dx\]

Evaluate of the following integral:

\[\int 3^{2 \log_3} {}^x dx\]

Evaluate of the following integral:

\[\int \log_x \text{x  dx}\] 

Evaluate: 

\[\int\sqrt{\frac{1 + \cos 2x}{2}}dx\]

Evaluate:

\[\int\sqrt{\frac{1 - \cos 2x}{2}}dx\]

Evaluate : 

\[\int\frac{e^{6 \log_e x} - e^{5 \log_e x}}{e^{4 \log_e x} - e^{3 \log_e x}}dx\]

Evaluate: 

\[\int\frac{1}{a^x b^x}dx\]

Evaluate: 

\[\int\frac{2 \cos^2 x - \cos 2x}{\cos^2 x}dx\]

Evaluate the following definite integral:

\[\int_0^1 \frac{1}{\sqrt{\left( x - 1 \right)\left( 2 - x \right)}}dx\]

Evaluate the following integral:

\[\int\limits_1^2 \left| x - 3 \right| dx\]

Evaluate the following integral:

\[\int\limits_0^4 \left| x - 1 \right| dx\]

Evaluate the following integral:

\[\int\limits_0^4 \left( \left| x \right| + \left| x - 2 \right| + \left| x - 4 \right| \right) dx\]

Evaluate each of the following integral:

\[\int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{\tan^2 x}{1 + e^x}dx\]

 


Evaluate each of the following integral:

\[\int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{x^{11} - 3 x^9 + 5 x^7 - x^5 + 1}{\cos^2 x}dx\]

Evaluate the following integral:

\[\int_0^\frac{\pi}{2} \frac{\tan^7 x}{\tan^7 x + \cot^7 x}dx\]

Evaluate the following integral:

\[\int_2^8 \frac{\sqrt{10 - x}}{\sqrt{x} + \sqrt{10 - x}}dx\]

Evaluate the following integral:

\[\int_{- \frac{3\pi}{2}}^{- \frac{\pi}{2}} \left\{ \sin^2 \left( 3\pi + x \right) + \left( \pi + x \right)^3 \right\}dx\]

Evaluate the following integral:

\[\int_0^\pi \left( \frac{x}{1 + \sin^2 x} + \cos^7 x \right)dx\]

Evaluate the following integral:

\[\int_0^\frac{\pi}{2} \frac{a\sin x + b\sin x}{\sin x + \cos x}dx\]

 


Evaluate:  `int_-1^2 (|"x"|)/"x"d"x"`.


Evaluate: `int_1^5{|"x"-1|+|"x"-2|+|"x"-3|}d"x"`.


`int_(pi/5)^((3pi)/10) [(tan x)/(tan x + cot x)]`dx = ?


Find: `int (dx)/sqrt(3 - 2x - x^2)`


The value of `int_0^1 (x^4(1 - x)^4)/(1 + x^2) dx` is


Evaluate: `int x/(x^2 + 1)"d"x`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×