Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int_0^\frac{\pi}{2} \sqrt{\sin \phi} \cos^5 \phi\ d \phi\]
\[Let\ \sin \phi = t . Then, \cos \phi\ d\phi = dt\]
\[When\ \phi = 0, t = 0\ and\ \phi = \frac{\pi}{2}, t = 1\]
\[Also, \cos^5 \phi = \cos^4 \phi \cos \phi = \left( 1 - \sin^2 \phi \right)^2 \cos \phi\]
\[ \therefore I = \int_0^\frac{\pi}{2} \sqrt{\sin \phi} \cos^5 \phi d \phi\]
\[ \Rightarrow I = \int_0^1 \sqrt{t} \left( 1 - t^2 \right)^2 dt\]
\[ \Rightarrow I = \int_0^1 \sqrt{t}\left( 1 + t^4 - 2 t^2 \right) dt\]
\[ \Rightarrow I = \int_0^1 \left( \sqrt{t} + t^\frac{9}{2} - 2 t^\frac{5}{2} \right) dt\]
\[ \Rightarrow I = \left[ \frac{2 t^\frac{3}{2}}{3} + \frac{2 t^\frac{11}{2}}{11} - \frac{4 t^\frac{7}{2}}{7} \right]_0^1 \]
\[ \Rightarrow I = \frac{2}{3} + \frac{2}{11} - \frac{4}{7}\]
\[ \Rightarrow I = \frac{64}{231}\]
APPEARS IN
संबंधित प्रश्न
Evaluate each of the following integral:
Evaluate :
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
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:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Choose the correct alternative:
Γ(n) is
Choose the correct alternative:
`Γ(3/2)`
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
