Advertisements
Advertisements
Question
Advertisements
Solution
\[Let\ I = \int_0^\pi \left( \sin^2 \frac{x}{2} - \cos^2 \frac{x}{2} \right) d\ x\ . Then, \]
\[I = - \int_0^\pi \cos\ x\ dx\ \left[ \because \cos A = \cos^2 \frac{A}{2} - \sin^2 \frac{A}{2} \right]\]
\[ \Rightarrow I = - \left[ \sin x \right]_0^\pi \]
\[ \Rightarrow I = 0\]
APPEARS IN
RELATED QUESTIONS
Evaluate the following integral:
Evaluate each of the following integral:
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
\[\int\limits_0^4 x dx\]
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Choose the correct alternative:
If n > 0, then Γ(n) is
Choose the correct alternative:
`Γ(3/2)`
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
The value of `int_2^3 x/(x^2 + 1)`dx is ______.
