English

Π / 4 ∫ 0 Sec X D X

Advertisements
Advertisements

Question

\[\int\limits_0^{\pi/4} \sec x dx\]
Advertisements

Solution

\[Let I = \int_0^\frac{\pi}{4} \sec x\ d\ x\ . Then, \]
\[I = \int_0^\frac{\pi}{4} \sec x \frac{\sec x + \tan x}{\sec x + \tan x} d x\]
\[ \Rightarrow I = \int_0^\frac{\pi}{4} \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} d x\]
\[Put u = \sec x + \tan x\]
\[ \Rightarrow du = \sec^2 x + \sec x \tan x dx\]
\[ \therefore \int_0^\frac{\pi}{4} \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} d x = \int\frac{du}{u}\]
\[ \Rightarrow I = \left[ \log u \right]\]
\[ \Rightarrow I = \left[ \log \left( \sec x + \tan x \right) \right]_0^\frac{\pi}{4} \]
\[ \Rightarrow I = \log \left( \sec\frac{\pi}{4} + \tan\frac{\pi}{4} \right) - \log \left( \sec 0 + \tan 0 \right)\]
\[ \Rightarrow I = \log (\sqrt{2} + 1) - \log 1\]
\[ \Rightarrow I = \log (\sqrt{2} + 1)\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Definite Integrals - Exercise 20.1 [Page 16]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Exercise 20.1 | Q 12 | Page 16

RELATED QUESTIONS

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

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

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

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

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

\[\int_0^\frac{\pi}{4} \left( a^2 \cos^2 x + b^2 \sin^2 x \right)dx\]

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

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

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

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

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

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

\[\int\limits_3^5 \left( 2 - x \right) dx\]

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

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

\[\int\limits_0^3 \frac{1}{x^2 + 9} dx .\]

Evaluate each of the following integral:

\[\int_e^{e^2} \frac{1}{x\log x}dx\]

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

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

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

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

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

The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is

 


The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is 

 


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


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


\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]


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


\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]


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


Using second fundamental theorem, evaluate the following:

`int_0^(pi/2) sqrt(1 + cos x)  "d"x`


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_(- pi/2)^(pi/2) sin^2theta  "d"theta`


Evaluate the following:

`int_0^oo "e"^(- x/2) x^5  "d"x`


Choose the correct alternative:

If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x)  "d"x + int_"c"^"b" f(x)  "d"x` is


Choose the correct alternative:

Using the factorial representation of the gamma function, which of the following is the solution for the gamma function Γ(n) when n = 8 is


Choose the correct alternative:

If n > 0, then Γ(n) is


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


If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.


Verify the following:

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×