Advertisements
Advertisements
Question
Evaluate the following integral:
Advertisements
Solution
\[\int_0^\frac{\pi}{2} \left| \cos 2x \right| d x\]
\[\text{We know that}, \left| \cos 2x \right| = \begin{cases} - \cos 2x &,& \frac{\pi}{4} \leq x \leq \frac{\pi}{2}\\\cos 2x&,& 0 < x \leq \frac{\pi}{4}\end{cases}\]
\[ \therefore I = \int_{- 2}^2 \left| \cos 2x \right| d x\]
\[ \Rightarrow I = \int_0^\frac{\pi}{4} \cos 2x dx - \int_\frac{\pi}{4}^\frac{\pi}{2} \cos 2x dx\]
\[ \Rightarrow I = \left[ \frac{\sin 2x}{2} \right]_0^\frac{\pi}{4} - \left[ \frac{\sin 2x}{2} \right]_\frac{\pi}{4}^\frac{\pi}{2} \]
\[ \Rightarrow I = \frac{1}{2} - 0 - 0 + \frac{1}{2}\]
\[ \Rightarrow I = 1\]
APPEARS IN
RELATED QUESTIONS
Evaluate :`int_0^(pi/2)1/(1+cosx)dx`
Evaluate : `int_0^4(|x|+|x-2|+|x-4|)dx`
Evaluate : `int1/(3+5cosx)dx`
Evaluate :
`∫_(-pi)^pi (cos ax−sin bx)^2 dx`
find `∫_2^4 x/(x^2 + 1)dx`
Evaluate the integral by using substitution.
`int_1^2 (1/x- 1/(2x^2))e^(2x) dx`
Evaluate of the following integral:
Evaluate of the following integral:
Evaluate:
Evaluate:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate each of the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate
\[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\]
Find : \[\int e^{2x} \sin \left( 3x + 1 \right) dx\] .
Evaluate: `int_-1^2 (|"x"|)/"x"d"x"`.
Evaluate: `int_1^5{|"x"-1|+|"x"-2|+|"x"-3|}d"x"`.
If `I_n = int_0^(pi/4) tan^n theta "d"theta " then " I_8 + I_6` equals ______.
`int_0^1 x(1 - x)^5 "dx" =` ______.
Evaluate: `int_0^(π/2) sin 2x tan^-1 (sin x) dx`.
In Method: Resubstitution, what should be done first?
Which limits are used to evaluate the resulting expression in Method: Resubstitution?
Under Method: Changing the Limits, what does the lower limit \[a\] become?
After changing the limits to the new variable, how is the definite integral completed?
After using \[t=\tan^{-1}x\] in \[\int_{0}^{1}\frac{\tan^{-1}x}{1+x^2}\,dx\], which integral results?
