मराठी

Π ∫ 0 1 3 + 2 Sin X + Cos X D X

Advertisements
Advertisements

प्रश्न

\[\int\limits_0^\pi \frac{1}{3 + 2 \sin x + \cos x} dx\]
Advertisements

उत्तर

\[Let\ I = \int_0^\pi \frac{1}{3 + 2 \sin x + \cos x} d x . Then, \]
\[I = \int_0^\pi \frac{1}{3 + 2\left( \frac{2 \tan\frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right) + \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}} d x\]
\[ \Rightarrow I = \int_0^\pi \frac{1 + \tan^2 \frac{x}{2}}{2 \tan^2 \frac{x}{2} + 4 \tan \frac{x}{2} + 4} dx\]
\[Let \tan \frac{x}{2} = t . Then, \frac{1}{2} \sec^2 \frac{x}{2} dx = dt\]
\[When\ x = 0, t = 0\ and\ x = \pi, t = \infty \]
\[ \therefore I = \int_0^\infty \frac{2 dt}{2 t^2 + 4t + 4}\]
\[ \Rightarrow I = \int_0^\infty \frac{dt}{\left( t + 1 \right)^2 + 1}\]
\[ \Rightarrow I = \left[ \tan^{- 1} \left( t + 1 \right) \right]_0^\infty \]
\[ \Rightarrow I = \frac{\pi}{2} - \frac{\pi}{4}\]
\[ \Rightarrow I = \frac{\pi}{4}\]

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 22 | पृष्ठ ३९

संबंधित प्रश्‍न

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

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

\[\int\limits_0^1 \frac{2x + 3}{5 x^2 + 1} dx\]

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

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

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

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

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

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

\[\int\limits_0^{\pi/4} \frac{\tan^3 x}{1 + \cos 2x} dx\]

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

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

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

\[\int\limits_1^4 f\left( x \right) dx, where\ f\left( x \right) = \begin{cases}4x + 3 & , & \text{if }1 \leq x \leq 2 \\3x + 5 & , & \text{if }2 \leq x \leq 4\end{cases}\]

 


Evaluate the following integral:

\[\int\limits_{- 2}^2 \left| 2x + 3 \right| dx\]

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

 


\[\int_0^1 | x\sin \pi x | dx\]

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

\[\int\limits_1^2 \left( x^2 - 1 \right) dx\]

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

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

\[\int\limits_{- 2}^1 \frac{\left| x \right|}{x} dx .\]

\[\int\limits_0^\infty e^{- x} dx .\]

\[\int\limits_0^\pi \cos^5 x\ dx .\]

If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.

 

 


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


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

If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals


Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .


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


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


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


Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`


Using second fundamental theorem, evaluate the following:

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


Evaluate the following using properties of definite integral:

`int_0^1 x/((1 - x)^(3/4))  "d"x`


Choose the correct alternative:

Γ(n) is


Verify the following:

`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×