Karnataka Board PUCPUC Science 2nd PUC Class 12

Evaluate : ∫30 dx/(9+x2) - Mathematics

Advertisements
Advertisements

Evaluate : `int_0^3dx/(9+x^2)`

Advertisements

Solution

Given,

`I=int_0^3dx/(9+x^2)=int_0^3dx/(3^2+x^2)`

We know that, `intdx/(x^2+a^2)=1/atan^(-1)(x/a)+C`

Therefore,

`I=int_0^3dx/(x^2+3^2)`

`=1/3[tan^(-1)(x/3)]_0^3`

`=1/3[tan^(-1)1-tan^(-1)0]`

`=1/3[pi/4-0]`

`=pi/12`

 

  Is there an error in this question or solution?
2013-2014 (March) Delhi Set 1

RELATED QUESTIONS

Integrate the following w.r.t. x `(x^3-3x+1)/sqrt(1-x^2)`


`∫   x    \sqrt{x + 2}     dx ` 

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

\[\int\frac{1 + \tan x}{1 - \tan x} dx\]

\[\int\frac{1}{x \log x} dx\]

\[\int\frac{1}{e^x + 1} dx\]

` ∫ {cot x}/ { log sin x} dx `

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

\[\int\frac{sec x}{\log \left( \text{sec x }+ \text{tan x} \right)} dx\]

\[\int\frac{10 x^9 + {10}^x \log_e 10}{{10}^x + x^{10}} dx\]

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


Evaluate the following integrals:

\[\int\frac{\sqrt{1 + x^2}}{x^4}dx\]

Evaluate the following integrals:

\[\int\frac{1}{\left( x^2 + 2x + 10 \right)^2}dx\]

 


`  ∫    {1} / {cos x  + "cosec x" } dx  `

Evaluate the following integrals:

\[\int\frac{5x - 2}{1 + 2x + 3 x^2} \text{ dx }\]

Evaluate the following integrals: 

\[\int\frac{x + 2}{\sqrt{x^2 + 2x + 3}} \text{ dx }\]

\[\int\frac{1}{5 - 4 \cos x} \text{ dx }\]

Evaluate the following integrals:

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

 


Evaluate the following integrals:

\[\int e^{2x} \text{ sin }\left( 3x + 1 \right) \text{ dx }\]

\[\int\left( x - 3 \right)\sqrt{x^2 + 3x - 18} \text{  dx }\]

Evaluate the following integral:

\[\int\frac{x^2 + 1}{\left( x^2 + 4 \right)\left( x^2 + 25 \right)}dx\]

Evaluate the following integral:

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

\[\int\frac{2x + 1}{\left( x + 2 \right) \left( x - 3 \right)^2} dx\]

Evaluate the following integral:

\[\int\frac{1}{x\left( x^3 + 8 \right)}dx\]

 


Evaluate the following integral:

\[\int\frac{2 x^2 + 1}{x^2 \left( x^2 + 4 \right)}dx\]

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

Evaluate the following integrals:

\[\int\frac{x^2}{(x - 1) ( x^2 + 1)}dx\]

Evaluate the following integral:

\[\int\frac{x^2}{x^4 + x^2 - 2}dx\]

\[\int\frac{x^2 + 1}{x^4 - x^2 + 1} \text{ dx }\]

Evaluate the following integral:

\[\int\frac{1}{\sin^4 x + \sin^2 x \cos^2 x + \cos^4 x}dx\]

Evaluate:\[\int\frac{\sin \sqrt{x}}{\sqrt{x}} \text{ dx }\]


Evaluate:\[\int \sec^2 \left( 7 - 4x \right) \text{ dx }\]


Evaluate: \[\int\frac{1}{\sqrt{1 - x^2}} \text{ dx }\]


Evaluate: \[\int\left( 1 - x \right)\sqrt{x}\text{  dx }\]


Evaluate:

\[\int \cos^{-1} \left(\sin x \right) \text{dx}\]


Evaluate: `int_  (x + sin x)/(1 + cos x )  dx`


Evaluate the following:

`int sqrt(5 - 2x + x^2) "d"x`


Evaluate the following:

`int x/(x^4 - 1) "d"x`


Evaluate the following:

`int ("d"x)/(xsqrt(x^4 - 1))`  (Hint: Put x2 = sec θ)


Evaluate the following:

`int_1^2 ("d"x)/sqrt((x - 1)(2 - x))`


Share
Notifications



      Forgot password?
Use app×