English

Π ∫ 0 1 3 + 2 Sin X + Cos X D X

Advertisements
Advertisements

Question

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

Solution

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

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Definite Integrals - Exercise 20.2 [Page 39]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Exercise 20.2 | Q 22 | Page 39

RELATED QUESTIONS

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

\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]


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

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

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

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

\[\int_\frac{\pi}{6}^\frac{\pi}{3} \left( \tan x + \cot x \right)^2 dx\]

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

\[\int\limits_0^\pi 5 \left( 5 - 4 \cos \theta \right)^{1/4} \sin \theta\ d \theta\]

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

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

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

\[\int_0^{2\pi} \cos^{- 1} \left( \cos x \right)dx\]

\[\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x} + \sqrt[3]{7} - x} dx\]

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

 


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

\[\int\limits_2^3 \left( 2 x^2 + 1 \right) dx\]

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

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

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

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

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

If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.

 


Write the coefficient abc of which the value of the integral

\[\int\limits_{- 3}^3 \left( a x^2 + bx + c \right) dx\] is independent.

The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is

 


The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .


The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is 


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


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


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


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


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


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


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


\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]


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")`


Evaluate the following:

`int_0^oo "e"^(-4x) x^4  "d"x`


Evaluate the following:

`int_0^oo "e"^(- x/2) x^5  "d"x`


Choose the correct alternative:

`int_0^oo "e"^(-2x)  "d"x` is


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×