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
Evaluate the following integral:
If f(2a − x) = −f(x), prove that
Evaluate each of the following integral:
Solve each of the following integral:
Write the coefficient a, b, c of which the value of the integral
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.
