हिंदी

Π / 2 ∫ 0 X Sin 2 X + Cos 2 X D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

योग

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
Definite Integrals
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 20: Definite Integrals - Revision Exercise [पृष्ठ १२२]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Revision Exercise | Q 54 | पृष्ठ १२२

संबंधित प्रश्न

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

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

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

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

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

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

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

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

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

\[\int\limits_{- 1}^1 5 x^4 \sqrt{x^5 + 1} dx\]

\[\int_0^\frac{1}{2} \frac{1}{\left( 1 + x^2 \right)\sqrt{1 - x^2}}dx\]

Evaluate the following integral:

\[\int\limits_{- 3}^3 \left| x + 1 \right| dx\]

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

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

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

If f(2a − x) = −f(x), prove that

\[\int\limits_0^{2a} f\left( x \right) dx = 0 .\]

If f(x) is a continuous function defined on [−aa], then prove that 

\[\int\limits_{- a}^a f\left( x \right) dx = \int\limits_0^a \left\{ f\left( x \right) + f\left( - x \right) \right\} dx\]

\[\int\limits_0^2 \left( x + 3 \right) dx\]

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

\[\int\limits_0^1 \left( 3 x^2 + 5x \right) dx\]

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

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

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

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

\[\int\limits_1^2 \log_e \left[ x \right] dx .\]

\[\int\limits_0^{\pi/2} \frac{\cos x}{\left( 2 + \sin x \right)\left( 1 + \sin x \right)} dx\] equals

\[\int\limits_{- 1}^1 \left| 1 - x \right| dx\]  is equal to

Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .


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


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


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


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


\[\int\limits_0^\pi \cos 2x \log \sin x dx\]


\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]


Using second fundamental theorem, evaluate the following:

`int_0^(1/4) sqrt(1 - 4)  "d"x`


Using second fundamental theorem, evaluate the following:

`int_1^2 (x "d"x)/(x^2 + 1)`


Evaluate the following:

f(x) = `{{:("c"x",", 0 < x < 1),(0",",  "otherwise"):}` Find 'c" if `int_0^1 "f"(x)  "d"x` = 2


Evaluate the following using properties of definite integral:

`int_0^1 x/((1 - x)^(3/4))  "d"x`


Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`


Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×