English

Π / 2 ∫ 0 √ Sin ϕ Cos 5 ϕ D ϕ

Advertisements
Advertisements

Question

\[\int\limits_0^{\pi/2} \sqrt{\sin \phi} \cos^5 \phi\ d\phi\]

 

Advertisements

Solution

\[\int_0^\frac{\pi}{2} \sqrt{\sin \phi} \cos^5 \phi\ d \phi\]

\[Let\ \sin \phi = t . Then, \cos \phi\ d\phi = dt\]

\[When\ \phi = 0, t = 0\ and\ \phi = \frac{\pi}{2}, t = 1\]

\[Also, \cos^5 \phi = \cos^4 \phi \cos \phi = \left( 1 - \sin^2 \phi \right)^2 \cos \phi\]

\[ \therefore I = \int_0^\frac{\pi}{2} \sqrt{\sin \phi} \cos^5 \phi d \phi\]

\[ \Rightarrow I = \int_0^1 \sqrt{t} \left( 1 - t^2 \right)^2 dt\]

\[ \Rightarrow I = \int_0^1 \sqrt{t}\left( 1 + t^4 - 2 t^2 \right) dt\]

\[ \Rightarrow I = \int_0^1 \left( \sqrt{t} + t^\frac{9}{2} - 2 t^\frac{5}{2} \right) dt\]

\[ \Rightarrow I = \left[ \frac{2 t^\frac{3}{2}}{3} + \frac{2 t^\frac{11}{2}}{11} - \frac{4 t^\frac{7}{2}}{7} \right]_0^1 \]

\[ \Rightarrow I = \frac{2}{3} + \frac{2}{11} - \frac{4}{7}\]

\[ \Rightarrow I = \frac{64}{231}\]

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 11 | Page 39

RELATED QUESTIONS

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

\[\int\limits_{\pi/3}^{\pi/4} \left( \tan x + \cot x \right)^2 dx\]

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

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

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

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

\[\int\limits_{- 1}^1 \frac{1}{x^2 + 2x + 5} dx\]

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

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

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

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

\[\int\limits_0^9 f\left( x \right) dx, where f\left( x \right) \begin{cases}\sin x & , & 0 \leq x \leq \pi/2 \\ 1 & , & \pi/2 \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9\end{cases}\]

\[\int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx\]

 


Evaluate each of the following integral:

\[\int_0^{2\pi} \log\left( \sec x + \tan x \right)dx\]

 


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

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

 


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

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

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

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

\[\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{a - \sin \theta}{a + \sin \theta} \right) d\theta\]

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

 

 


If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.


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

The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]

 


\[\int\limits_0^1 \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\} dx\] is equal to

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


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


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


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


\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]


\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]


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 integrals as the limit of the sum:

`int_0^1 x^2  "d"x`


Choose the correct alternative:

Γ(1) is


Find `int x^2/(x^4 + 3x^2 + 2) "d"x`


Find `int sqrt(10 - 4x + 4x^2)  "d"x`


Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×