Advertisements
Advertisements
Question
Advertisements
Solution
\[Let\ I = \int_0^\frac{\pi}{2} x^2 \cos x d x . Then, \]
\[\text{Integrating by parts}\]
\[I = \left[ x^2 \sin x \right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} 2x \sin x d x\]
\[ \Rightarrow I = \left[ x^2 \sin x \right]_0^\frac{\pi}{2} - \left[ - 2x \cos x \right]_0^\frac{\pi}{2} + \int_0^\frac{\pi}{2} - 2 \cos x d x\]
\[ \Rightarrow I = \left[ x^2 \sin x \right]_0^\frac{\pi}{2} + \left[ 2x \cos x \right]_0^\frac{\pi}{2} - \left[ 2 \sin x \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = \frac{\pi^2}{4} - 2\]
APPEARS IN
RELATED QUESTIONS
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
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_0^(pi/2) sqrt(1 + cos x) "d"x`
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
