हिंदी

Evaluate the following: d∫dx16-9x2 - Mathematics

Advertisements
Advertisements

प्रश्न

Evaluate the following:

`int ("d"x)/sqrt(16 - 9x^2)`

योग

उत्तर

Let I = `int ("d"x)/sqrt(16 - 9x^2)`

= `1/3 int ("d"x)/sqrt(16/9 - x^2)`

= `1/3 int ("d"x)/sqrt((4/3)^2 - x^2)`

= `1/3 sin^-1  x/(4/3) + "C"`  ....`[because int ("d"x)/sqrt("a"^2 - x^2) = sin^-1  x/"a" + "C"]`

∴ I = `1/3 sin^-1  (3x)/4 + "C"`

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

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 7 Integrals
Exercise | Q 14 | पृष्ठ १६४

संबंधित प्रश्न

`∫   x    \sqrt{x + 2}     dx ` 

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

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

\[\int\frac{2 \cos x - 3 \sin x}{6 \cos x + 4 \sin x} dx\]

\[\int\frac{1}{\cos 3x - \cos x} dx\]

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

Evaluate the following integrals:

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

 


Evaluate the following integrals:

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

Evaluate the following integrals: 

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

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

\[\int e^{2x} \text{ sin x cos x dx }\]

Evaluate the following integrals:

\[\int\left( x + 3 \right)\sqrt{3 - 4x - x^2} \text{  dx }\]

\[\int(3x + 1) \sqrt{4 - 3x - 2 x^2} \text{  dx }\]

Evaluate the following integral :-

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

Evaluate the following integral:

\[\int\frac{x^2}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)}dx\]

Evaluate the following integral:

\[\int\frac{x^2}{x^4 - x^2 - 12}dx\]

 


Evaluate the following integral:

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

\[\int\frac{( x^2 + 1) ( x^2 + 4)}{( x^2 + 3) ( x^2 - 5)} dx\]

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


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


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


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


Evaluate the following:

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

Englishहिंदीमराठी


      Forgot password?
Use app×