Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_\frac{\pi}{6}^\frac{\pi}{4} cosec x d x . Then, \]
\[I = \int_\frac{\pi}{6}^\frac{\pi}{4} cosec\ x \frac{cosec\ x - \cot x}{cosec x - \cot x} d x\]
\[ \Rightarrow I = \int_\frac{\pi}{6}^\frac{\pi}{4} \frac{{cosec}^2\ x - cosec\ x \cot x}{cosec\ x\ - \cot x} d x\]
\[ \Rightarrow I = \left[ \log \left( cosec\ x - \cot x \right) \right]_\frac{\pi}{6}^\frac{\pi}{4} \]
\[ \Rightarrow I = \log \left( \sqrt{2} - 1 \right) - \log\left( 2 - \sqrt{3} \right)\]
APPEARS IN
संबंधित प्रश्न
Evaluate each of the following integral:
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
Evaluate the following:
f(x) = `{{:("c"x",", 0 < x < 1),(0",", "otherwise"):}` Find 'c" if `int_0^1 "f"(x) "d"x` = 2
Evaluate the following using properties of definite integral:
`int_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
Choose the correct alternative:
`int_0^1 (2x + 1) "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
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Which integral has lower and upper limits?
What are definite integrals used to find over a fixed interval?
