मराठी

Evaluate the integral by using substitution. ∫0π2sinx1+cos2xdx - Mathematics

Advertisements
Advertisements

प्रश्न

Evaluate the integral by using substitution.

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

बेरीज

उत्तर

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

Substituting cos x = t,

⇒ - sin x dx = dt

And x = 0, t = 1, x `= pi/2,` t = 0

Hence, `I = - int_1^0 1/(1 + t^2)`  dt

`= - [tan^-1 t]_1^0`

`= - [tan^-1 0 - tan^-1 1]`

`= - [0 - pi/4]`

`= pi/4`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Integrals - Exercise 7.10 [पृष्ठ ३४०]

APPEARS IN

एनसीईआरटी Mathematics [English] Class 12
पाठ 7 Integrals
Exercise 7.10 | Q 5 | पृष्ठ ३४०

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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


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


 

Evaluate `int_(-1)^2|x^3-x|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_0^1 sin^(-1) ((2x)/(1+ x^2)) dx`


Evaluate of the following integral: 

\[\int x^\frac{5}{4} dx\]

Evaluate of the following integral:

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

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 : 

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

Evaluate: 

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

Evaluate:

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

Evaluate the following integral:

\[\int\limits_0^3 \left| 3x - 1 \right| dx\]

 


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:

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

 


Evaluate each of the following integral:

\[\int_{- a}^a \frac{1}{1 + a^x}dx\]`, a > 0`

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:

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

Evaluate the following integral:

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

Evaluate the following integral:

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

Evaluate the following integral:

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

Evaluate the following integral:

\[\int_0^{2\pi} \sin^{100} x \cos^{101} xdx\]

 


Evaluate the following integral:

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

 


Evaluate : \[\int\limits_{- 2}^1 \left| x^3 - x \right|dx\] .


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


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


Evaluate: `int_  e^x ((2+sin2x))/cos^2 x dx`


Evaluate: `int_-π^π (1 - "x"^2) sin "x" cos^2 "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 = ?


Evaluate the following:

`int ("e"^(6logx) - "e"^(5logx))/("e"^(4logx) - "e"^(3logx)) "d"x`


Evaluate the following:

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


If `int x^5 cos (x^6)"d"x = "k" sin (x^6) + "C"`, find the value of k.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×