English

Π / 4 ∫ 0 Tan 3 X 1 + Cos 2 X D X - Mathematics

Advertisements
Advertisements

Question

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

Solution

\[Let\ I = \int_0^\frac{\pi}{4} \frac{\tan^3 x}{1 + \cos 2x} d\ x . Then, \]
\[I = \int_0^\frac{\pi}{4} \frac{\tan^3 x}{2 \cos^2 x} d\ x\]
\[ \Rightarrow I = \frac{1}{2} \int_0^\frac{\pi}{4} \tan^3 x \sec^2 x dx\]
\[Let \tan\ x = t . Then, \sec^2 x\ dx\ = dt\]
\[When\ x = 0, t = 0\ and\ x\ = \frac{\pi}{4}, t = 1\]
\[ \therefore I = \frac{1}{2} \int_0^1 t^3 dt\]
\[ \Rightarrow I = \frac{1}{2} \left[ \frac{t^4}{4} \right]_0^1 \]
\[ \Rightarrow I = \frac{1}{2}\left( \frac{1}{4} - 0 \right)\]
\[ \Rightarrow I = \frac{1}{8}\]

shaalaa.com
Definite Integrals
  Is there an error in this question or solution?
Chapter 20: Definite Integrals - Exercise 20.2 [Page 39]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 20 Definite Integrals
Exercise 20.2 | Q 26 | Page 39

RELATED QUESTIONS

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

\[\int\limits_1^3 \frac{\log x}{\left( x + 1 \right)^2} dx\]

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

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

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

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

\[\int_0^\frac{\pi}{4} \left( \tan x + \cot x \right)^{- 2} dx\]

\[\int\limits_2^4 \frac{x}{x^2 + 1} dx\]

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

\[\int\limits_1^3 \frac{\cos \left( \log x \right)}{x} dx\]

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

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

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

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

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

 


Evaluate each of the following integral:

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

 


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

 


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

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

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

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

\[\int\limits_0^4 \frac{1}{\sqrt{16 - x^2}} dx .\]

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

\[\int\limits_a^b \frac{f\left( x \right)}{f\left( x \right) + f\left( a + b - x \right)} dx .\]

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

Evaluate : 

\[\int\limits_2^3 3^x dx .\]

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

 


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

Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .


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


Using second fundamental theorem, evaluate the following:

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


Using second fundamental theorem, evaluate the following:

`int_0^(pi/2) sqrt(1 + cos x)  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_1^3 x  "d"x`


Choose the correct alternative:

If n > 0, then Γ(n) is


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


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


Evaluate the following:

`int ((x^2 + 2))/(x + 1) "d"x`


If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.


Find: `int logx/(1 + log x)^2 dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×