English

Π / 4 ∫ 0 Tan 2 X D X . - Mathematics

Advertisements
Advertisements

Question

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

Solution

\[\int_0^\frac{\pi}{4} \tan^2 x\ d x\]

\[ = \int_0^\frac{\pi}{4} \left( se c^2 x - 1 \right) d x\]

\[ = \left[ \tan x - x \right]_0^\frac{\pi}{4} \]

\[ = 1 - \frac{\pi}{4} - 0\]

\[ = 1 - \frac{\pi}{4}\]

shaalaa.com
Definite Integrals
  Is there an error in this question or solution?
Chapter 20: Definite Integrals - Very Short Answers [Page 115]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 20 Definite Integrals
Very Short Answers | Q 7 | Page 115

RELATED QUESTIONS

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

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

 


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

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

\[\int_0^1 x\log\left( 1 + 2x \right)dx\]

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

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

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

\[\int\limits_0^{\pi/4} \left( \sqrt{\tan}x + \sqrt{\cot}x \right) dx\]

\[\int_0^\frac{\pi}{2} \frac{\cos^2 x}{1 + 3 \sin^2 x}dx\]

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

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

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

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

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

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

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

 


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

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

\[\int\limits_0^{\pi/2} \sqrt{1 - \cos 2x}\ dx .\]

Evaluate each of the following  integral:

\[\int_0^1 x e^{x^2} dx\]

 


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


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

The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is 


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


\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\cot}x} dx\] is

\[\int\limits_1^e \log x\ dx =\]

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

 


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


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


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


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


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


Using second fundamental theorem, evaluate the following:

`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`


Evaluate the following:

f(x) = `{{:("c"x",", 0 < x < 1),(0",",  "otherwise"):}` Find 'c" if `int_0^1 "f"(x)  "d"x` = 2


Choose the correct alternative:

`int_(-1)^1 x^3 "e"^(x^4)  "d"x` is


Choose the correct alternative:

The value of `int_(- pi/2)^(pi/2) cos  x  "d"x` is


Choose the correct alternative:

`int_0^oo x^4"e"^-x  "d"x` is


Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x


Verify the following:

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


`int (x + 3)/(x + 4)^2 "e"^x  "d"x` = ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×