Advertisements
Advertisements
Question
Advertisements
Solution
\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \cos^2 x\ d x\]
\[ = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{1 + \cos2x}{2} dx\]
\[ = \frac{1}{2} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 1 + \cos2x \right) dx\]
\[ = \frac{1}{2} \left[ x + \frac{\sin2x}{2} \right]_{- \frac{\pi}{2}}^\frac{\pi}{2} \]
\[ = \frac{1}{2}\left( \frac{\pi}{2} + 0 + \frac{\pi}{2} - 0 \right)\]
\[ = \frac{\pi}{2}\]
APPEARS IN
RELATED QUESTIONS
Evaluate each of the following integral:
Evaluate each of the following integral:
\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]
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 \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
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_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Evaluate the following using properties of definite integral:
`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x) "d"x`
