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

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

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

 


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

\[\int\limits_1^e \frac{\log x}{x} dx\]

\[\int\limits_1^2 \frac{3x}{9 x^2 - 1} dx\]

\[\int_0^\frac{\pi}{4} \frac{\sin x + \cos x}{3 + \sin2x}dx\]

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

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

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

\[\int_\frac{1}{3}^1 \frac{\left( x - x^3 \right)^\frac{1}{3}}{x^4}dx\]

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

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

\[\int_{- 2}^2 x e^\left| x \right| dx\]

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

Evaluate each of the following integral:

\[\int_0^{2\pi} \log\left( \sec x + \tan x \right)dx\]

 


Evaluate each of the following integral:

\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]


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

 


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

\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx, 0 < \alpha < \pi\]

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

If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]


\[\int\limits_0^1 \sqrt{x \left( 1 - x \right)} dx\] equals

\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\]  equals


The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is

 


The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is 

 


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


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


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


\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]


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


\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]


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


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


Evaluate the following integrals as the limit of the sum:

`int_1^3 x  "d"x`


`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.


What is the result of a definite integral?


What is the meaning of an indefinite integral?


What is the meaning of a definite integral?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×