Advertisements
Advertisements
Question
Solution
\[Let\ I = \int_\frac{\pi}{6}^\frac{\pi}{4} cosec x d x . Then, \]
\[I = \int_\frac{\pi}{6}^\frac{\pi}{4} cosec\ x \frac{cosec\ x - \cot x}{cosec x - \cot x} d x\]
\[ \Rightarrow I = \int_\frac{\pi}{6}^\frac{\pi}{4} \frac{{cosec}^2\ x - cosec\ x \cot x}{cosec\ x\ - \cot x} d x\]
\[ \Rightarrow I = \left[ \log \left( cosec\ x - \cot x \right) \right]_\frac{\pi}{6}^\frac{\pi}{4} \]
\[ \Rightarrow I = \log \left( \sqrt{2} - 1 \right) - \log\left( 2 - \sqrt{3} \right)\]
APPEARS IN
RELATED QUESTIONS
\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]
Evaluate :
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Evaluate the following using properties of definite integral:
`int_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Find `int sqrt(10 - 4x + 4x^2) "d"x`
