Advertisements
Advertisements
प्रश्न
उत्तर
\[Let\ I = \int_0^\frac{\pi}{2} \cos^3 x\ d\ x . Then, \]
\[I = \int_0^\frac{\pi}{2} \cos^2 x \cos\ x\ d\ x\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \left( 1 - \sin^2 x \right) \cos x d x\]
\[Let u = \sin x, du = \cos\ x\ dx\]
\[ \Rightarrow I = \int\left( 1 - u^2 \right) du\]
\[ \Rightarrow I = \left[ u - \frac{u^3}{3} \right]\]
\[ \Rightarrow I = \left[ \sin x - \frac{\sin^3 x}{3} \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = 1 - \frac{1}{3} - 0\]
\[ \Rightarrow I = \frac{2}{3}\]
APPEARS IN
संबंधित प्रश्न
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
Evaluate the following:
`int_0^oo "e"^(-4x) x^4 "d"x`
Choose the correct alternative:
Γ(n) is
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
