मराठी

Π / 2 ∫ 0 Sin X Cos X 1 + Sin 4 X D X

Advertisements
Advertisements

प्रश्न

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

उत्तर

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

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Exercise 20.2 [पृष्ठ ३९]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Exercise 20.2 | Q 18 | पृष्ठ ३९

संबंधित प्रश्‍न

\[\int\limits_0^{\pi/2} \left( \sin x + \cos x \right) dx\]

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

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

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

\[\int\limits_0^1 \left( x e^{2x} + \sin\frac{\ pix}{2} \right) dx\]

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

\[\int\limits_2^4 \frac{x}{x^2 + 1} dx\]

\[\int\limits_1^2 \frac{1}{x \left( 1 + \log x \right)^2} dx\]

\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]

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

\[\int\limits_0^a x \sqrt{\frac{a^2 - x^2}{a^2 + x^2}} dx\]

\[\int_0^\frac{\pi}{2} \sqrt{\cos x - \cos^3 x}\left( \sec^2 x - 1 \right) \cos^2 xdx\]

\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{- \frac{\pi}{2}}{\sqrt{\cos x \sin^2 x}}dx\]

\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\tan x}} dx\]

\[\int\limits_0^\pi \frac{x \tan x}{\sec x \ cosec x} dx\]

\[\int\limits_1^2 x^2 dx\]

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

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

Solve each of the following integral:

\[\int_2^4 \frac{x}{x^2 + 1}dx\]

The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .


\[\int\limits_0^1 \tan^{- 1} x dx\]


\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]


\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]


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


\[\int\limits_0^1 \left| 2x - 1 \right| dx\]


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


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


\[\int\limits_1^4 \left( x^2 + x \right) dx\]


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


Evaluate the following using properties of definite integral:

`int_0^1 x/((1 - x)^(3/4))  "d"x`


Evaluate the following:

`Γ (9/2)`


Choose the correct alternative:

If n > 0, then Γ(n) is


Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`


If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.


Evaluate the following:

`int ((x^2 + 2))/(x + 1) "d"x`


`int x^9/(4x^2 + 1)^6  "d"x` is equal to ______.


Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`


The value of `int_2^3 x/(x^2 + 1)`dx is ______.


What are definite integrals used to find over a fixed interval?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×