मराठी

Π / 2 ∫ 0 1 1 + Cot 7 X D X

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

\[Let, I = \int_0^\frac{\pi}{2} \frac{1}{1 + co t^7 x} d x ..............(1)\]

\[ = \int_0^\frac{\pi}{2} \frac{1}{1 + co t^7 \left( \frac{\pi}{2} - x \right)} d x\]

\[ = \int_0^\frac{\pi}{2} \frac{1}{1 + \tan^7 x} d x ..............(2)\]

Adding (1) and (2)

\[2I = \int_0^\frac{\pi}{2} \frac{1}{1 + co t^7 x} + \frac{1}{1 + \tan^7 x} d x \]

\[ = \int_0^\frac{\pi}{2} \frac{2 + co t^7 x + \tan^7 x}{\left( 1 + co t^7 x \right)\left( 1 + \tan^7 x \right)}dx\]

\[ = \int_0^\frac{\pi}{2} \frac{2 + co t^7 x + \tan^7 x}{2 + co t^7 x + \tan^7 x}dx\]

\[ = \int_0^\frac{\pi}{2} dx\]

\[ = \left[ x \right]_0^\frac{\pi}{2} \]

\[ = \frac{\pi}{2}\]

\[Hence, I = \frac{\pi}{4}\]

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 37 | पृष्ठ १२२

संबंधित प्रश्‍न

\[\int\limits_{\pi/6}^{\pi/4} cosec\ x\ dx\]

\[\int\limits_{- \pi/4}^{\pi/4} \frac{1}{1 + \sin x} dx\]

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

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

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

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

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

\[\int\limits_0^\pi x \sin x \cos^4 x\ dx\]

\[\int\limits_0^\pi x \sin^3 x\ dx\]

If f is an integrable function, show that

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

 


If f (x) is a continuous function defined on [0, 2a]. Then, prove that

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

 


\[\int\limits_0^3 \left( x + 4 \right) dx\]

\[\int\limits_a^b \cos\ x\ dx\]

\[\int\limits_0^4 \left( x + e^{2x} \right) dx\]

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

\[\int\limits_{- \pi/2}^{\pi/2} \sin^3 x\ dx .\]

\[\int\limits_0^2 x\left[ x \right] dx .\]

If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:

\[\int\limits_0^{\pi/4} \sin \left\{ x \right\} dx\]

 


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

`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`


The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is 

 


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

\[\int\limits_0^{2a} f\left( x \right) dx\]  is equal to


Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .


\[\int\limits_1^2 x\sqrt{3x - 2} dx\]


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


\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]


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


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


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


Evaluate the following:

`int_0^2 "f"(x)  "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`


Evaluate the following using properties of definite integral:

`int_(-1)^1 log ((2 - x)/(2 + x))  "d"x`


Evaluate the following:

Γ(4)


Evaluate the following integrals as the limit of the sum:

`int_1^3 x  "d"x`


Choose the correct alternative:

`int_0^1 (2x + 1)  "d"x` is


Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×