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 the following definite integrals:
Prove that:
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Evaluate the following:
`int_0^2 "f"(x) "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
