Advertisements
Advertisements
Question
Advertisements
Solution
\[Let\ I = \int_1^3 \frac{\cos \left( \log x \right)}{x} d\ x . \]
\[Let\ \log\ x = t . Then, \frac{1}{x} dx = dt\]
\[When\ x = 1, t = 0\ and\ x\ = 3, t = \log 3\]
\[ \therefore I = \int_0^{\ log 3} \cos t d t\]
\[ = \left[ \sin t \right]_0^{\ log 3} \]
\[ = \sin \left( \log 3 \right)\]
APPEARS IN
RELATED QUESTIONS
Evaluate the following integral:
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_2^3 e^{- x} dx\]
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Choose the correct alternative:
Γ(1) is
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
What is an antiderivative of \[2x+1\]?
Evaluate \[\int_{1}^{3}(2x+1)\,dx\].
