Advertisements
Advertisements
Question
Advertisements
Solution
\[Let\ I = \int_0^\frac{\pi}{2} \cos^3 x\ d\ x . Then, \]
\[I = \int_0^\frac{\pi}{2} \cos^2 x \cos\ x\ d\ x\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \left( 1 - \sin^2 x \right) \cos x d x\]
\[Let u = \sin x, du = \cos\ x\ dx\]
\[ \Rightarrow I = \int\left( 1 - u^2 \right) du\]
\[ \Rightarrow I = \left[ u - \frac{u^3}{3} \right]\]
\[ \Rightarrow I = \left[ \sin x - \frac{\sin^3 x}{3} \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = 1 - \frac{1}{3} - 0\]
\[ \Rightarrow I = \frac{2}{3}\]
APPEARS IN
RELATED QUESTIONS
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
If f is an integrable function, show that
If f(x) is a continuous function defined on [−a, a], then prove that
Prove that:
Write the coefficient a, b, c of which the value of the integral
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
`int_0^(2a)f(x)dx`
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Choose the correct alternative:
If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x) "d"x + int_"c"^"b" f(x) "d"x` is
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
