Advertisements
Advertisements
Question
Solution
\[Let I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \sin^3 x\ d x\]
\[ = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \sin x \sin^2 x\ dx\]
\[ = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \sin x\left( 1 - \cos^2 x \right) dx\]
\[Let\ \cos x = t, then - \sin x\ dx = dt, \]
\[When\, x \to - \frac{\pi}{2} ; t \to 0\ and\ x \to \frac{\pi}{2} ; t \to 0\]
\[I = \int_0^0 \left( - 1 + t^2 \right) dt\]
\[\]
\[ = 0\]
APPEARS IN
RELATED QUESTIONS
\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]
Evaluate each of the following integral:
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Evaluate the following using properties of definite integral:
`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x) "d"x`
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
`int x^3/(x + 1)` is equal to ______.
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
Find: `int logx/(1 + log x)^2 dx`
