Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^\pi 5 \left( 5 - 4\cos \theta \right)^\frac{1}{4} \sin \theta\ d \theta . \]
\[Let\left( 5 - 4 \cos \theta \right) = t . Then, 4 \sin \theta\ d\theta = dt\]
\[When\ \theta = 0, t = 1\ and\ \theta = \pi, t = 9\]
\[ \therefore I = \frac{5}{4} \int_1^9 t^\frac{1}{4} dt\]
\[ \Rightarrow I = \frac{5}{4} \left[ \frac{4 t^\frac{5}{4}}{5} \right]_1^9 \]
\[ \Rightarrow I = \left( 9\sqrt{3} - 1 \right)\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
