हिंदी

Find: ∫ ( 3 X + 5 ) √ 5 + 4 X − 2 X 2 D X .

Advertisements
Advertisements

प्रश्न

Find: `int_  (3"x"+ 5)sqrt(5 + 4"x"-2"x"^2)d"x"`.

योग
Advertisements

उत्तर

`( 3x + 5)sqrt(5 + 4x - 2x^2) dx`

Let 3x + 5 = A(4 - 4x) + B

⇒ A = `-3/4, B = 8`

I = `3/4(4 - 4x)sqrt(5 + 4x + 2x^2) dx + 8 sqrt(5 + 4x - 2x^2)dx`

= `-3/4 I_1 + 8l_2 ("let")`

For I1, put 5 + 4x = - 2x2 = t

⇒ (4 - 4x) dx = dt

`-3/4 I_1 = - 3/4 sqrtt dt = - 3/4 xx 2/3 t^(3/2)`

= `-1/2 (5 + 4x - 2x^2 )^(3/2)`

`8I_2 = 8sqrt2 sqrt(7/2 - ( x - 1)^2) dx`

I = `-1/2 (5 + 4x - 2x^2 )^(3/2) + 4sqrt2(x - 1) sqrt(5/2 + 2x - x^2) + 14sqrt2 sin^-1  (sqrt2(x - 1))/sqrt7 + C`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2015-2016 (March) All India Set 1 E

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Evaluate : `int_0^4(|x|+|x-2|+|x-4|)dx`


Evaluate :

`int_e^(e^2) dx/(xlogx)`


Evaluate: `intsinsqrtx/sqrtxdx`

 


Evaluate the integral by using substitution.

`int_0^1 x/(x^2 +1)`dx


Evaluate the integral by using substitution.

`int_0^(pi/2) (sin x)/(1+ cos^2 x) dx`


Evaluate the integral by using substitution.

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


Evaluate of the following integral:

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

Evaluate: 

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

Evaluate:

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

Evaluate: 

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

Evaluate:

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

Evaluate: 

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

Evaluate:

\[\int\frac{e\log \sqrt{x}}{x}dx\]

Evaluate the following integral:

\[\int\limits_2^8 \left| x - 5 \right| dx\]

 


Evaluate the following integral:

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

Evaluate each of the following integral:

\[\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}dx\]

 


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:

\[\int_{- \frac{\pi}{3}}^\frac{\pi}{3} \frac{1}{1 + e^\ tan\ x}dx\]

 


Evaluate each of the following integral:

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

\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]

Evaluate the following integral:

\[\int_0^\pi x\sin x \cos^2 xdx\]

Evaluate the following integral:

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

Evaluate: \[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x}dx\] .


If `I_n = int_0^(pi/4) tan^n theta  "d"theta " then " I_8 + I_6` equals ______.


Evaluate the following:

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


Each student in a class of 40, studies at least one of the subjects English, Mathematics and Economics. 16 study English, 22 Economics and 26 Mathematics, 5 study English and Economics, 14 Mathematics and Economics and 2 study all the three subjects. The number of students who study English and Mathematics but not Economics is


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


After finding the antiderivative in Method: Resubstitution, what is the next step?


For the integral \[\int_{0}^{1}\frac{\tan^{-1}x}{1+x^2}\,dx\], which substitution is suitable?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×