हिंदी

Π / 4 ∫ 0 Sin 3 2 T Cos 2 T D T

Advertisements
Advertisements

प्रश्न

\[\int\limits_0^{\pi/4} \sin^3 2t \cos 2t\ dt\]
Advertisements

उत्तर

\[Let\ I = \int_0^\frac{\pi}{4} \sin^3 2t\ \cos 2t\ d\ t . Then, \]
\[Let\ \sin 2t = u . Then, 2 \cos\ 2t\ dt = du\]
\[When\ t = 0, u = 0\ and\ t\ = \frac{\pi}{4}, u = 1\]
\[ \therefore I = \frac{1}{2} \int_0^1 u^3 du\]
\[ \Rightarrow I = \frac{1}{2} \left[ \frac{u^4}{4} \right]_0^1 \]
\[ \Rightarrow I = \frac{1}{2}\left( \frac{1}{4} - 0 \right)\]
\[ \Rightarrow I = \frac{1}{8}\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - Exercise 20.2 [पृष्ठ ३९]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Exercise 20.2 | Q 41 | पृष्ठ ३९

संबंधित प्रश्न

\[\int\limits_0^1 \frac{1}{1 + x^2} dx\]

\[\int\limits_{\pi/3}^{\pi/4} \left( \tan x + \cot x \right)^2 dx\]

\[\int\limits_0^{\pi/2} \sqrt{1 + \sin x}\ dx\]

\[\int\limits_0^2 \frac{1}{4 + x - x^2} dx\]

\[\int_0^1 \frac{1}{1 + 2x + 2 x^2 + 2 x^3 + x^4}dx\]

\[\int\limits_0^1 \frac{2x}{1 + x^4} dx\]

\[\int\limits_0^{\pi/2} \frac{\sin x \cos x}{1 + \sin^4 x} dx\]

\[\int\limits_0^{\pi/2} \frac{1}{5 + 4 \sin x} dx\]

\[\int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx\]

 


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

\[\int_0^2 2x\left[ x \right]dx\]

\[\int\limits_0^\pi \frac{x \sin x}{1 + \sin x} dx\]

\[\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{2 - \sin x}{2 + \sin x} \right) dx\]

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

\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]

\[\int\limits_0^{\pi/2} \sqrt{1 - \cos 2x}\ dx .\]

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{2} e^x \left( \sin x - \cos x \right)dx\]

 


Evaluate : 

\[\int\limits_2^3 3^x dx .\]

\[\int\limits_0^3 \frac{3x + 1}{x^2 + 9} dx =\]

\[\lim_{n \to \infty} \left\{ \frac{1}{2n + 1} + \frac{1}{2n + 2} + . . . + \frac{1}{2n + n} \right\}\] is equal to

\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]


\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]


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


\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]


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


\[\int\limits_{- 1}^1 e^{2x} dx\]


\[\int\limits_2^3 e^{- x} dx\]


\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]


Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`


Using second fundamental theorem, evaluate the following:

`int_0^1 "e"^(2x)  "d"x`


Evaluate the following using properties of definite integral:

`int_(- pi/2)^(pi/2) sin^2theta  "d"theta`


Evaluate the following using properties of definite integral:

`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x)  "d"x`


Evaluate the following using properties of definite integral:

`int_0^1 log (1/x - 1)  "d"x`


Choose the correct alternative:

Γ(n) is


Choose the correct alternative:

`int_0^oo x^4"e"^-x  "d"x` is


Find `int x^2/(x^4 + 3x^2 + 2) "d"x`


Verify the following:

`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`


`int (x + 3)/(x + 4)^2 "e"^x  "d"x` = ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×