मराठी

Π / 2 ∫ 0 X Sin 2 X + Cos 2 X D X

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

\[Let, I = \int_0^\frac{\pi}{2} \frac{x}{\sin^2 x + \cos^2 x} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{x}{1} d x\]
\[ = \int_0^\frac{\pi}{2} x d x\]
\[ = \left[ \frac{x^2}{2} \right]_0^\frac{\pi}{2} \]
\[ \therefore I = \frac{\pi^2}{8}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Revision Exercise [पृष्ठ १२२]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Revision Exercise | Q 54 | पृष्ठ १२२

संबंधित प्रश्‍न

\[\int\limits_1^4 \frac{x^2 + x}{\sqrt{2x + 1}} dx\]

\[\int_0^\pi e^{2x} \cdot \sin\left( \frac{\pi}{4} + x \right) dx\]

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

\[\int\limits_0^{\pi/2} \sin^3 x\ dx\]

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

\[\int\limits_0^{\pi/2} \frac{dx}{a \cos x + b \sin x}a, b > 0\]

\[\int\limits_4^{12} x \left( x - 4 \right)^{1/3} dx\]

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

\[\int\limits_0^{\pi/6} \cos^{- 3} 2 \theta \sin 2\ \theta\ d\ \theta\]

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

\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{3/2}} dx\]

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

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

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

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

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


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


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


\[\int\limits_1^e \log x\ dx =\]

\[\int\limits_0^{\pi/2} x \sin x\ dx\]  is equal to

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


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


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


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


\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]


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


\[\int\limits_0^{15} \left[ x^2 \right] dx\]


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


Using second fundamental theorem, evaluate the following:

`int_1^"e" ("d"x)/(x(1 + logx)^3`


Using second fundamental theorem, evaluate the following:

`int_1^2 (x - 1)/x^2  "d"x`


Evaluate the following using properties of definite integral:

`int_(- pi/4)^(pi/4) x^3 cos^3 x  "d"x`


Evaluate the following using properties of definite integral:

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


Evaluate the following integrals as the limit of the sum:

`int_1^3 x  "d"x`


Choose the correct alternative:

`int_0^oo "e"^(-2x)  "d"x` is


Find `int sqrt(10 - 4x + 4x^2)  "d"x`


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 ______.


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


Using \[F\] as an antiderivative, how is \[\int_{a}^{b}f(x)\,dx\] evaluated?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×