Advertisements
Advertisements
Question
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Sum
Advertisements
Solution
We know cos 2x = `2cos^2x - 1`
⇒ cos x = `2cos^2 x/2 - 1`
⇒ 1 + cos x = `2cos^2 x/2`
`int_0^(pi/2) sqrt(2 cos^2 x/2) "d"x = int_0^(pi/2) sqrt(2) cos x/2 "d"x`
= `[(sqrt(2) sin x/2)/(1/2)]_0^(pi/2)`
= `2sqrt(2) sin pi/4 - 2sqrt(2) sin 0`
= `2sqrt(2) (1/sqrt(2))`
= 2
shaalaa.com
Definite Integrals
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\]
\[\int\limits_0^1 \frac{e^x}{1 + e^{2x}} dx\]
\[\int\limits_0^\pi 5 \left( 5 - 4 \cos \theta \right)^{1/4} \sin \theta\ d \theta\]
If f(x) is a continuous function defined on [−a, a], then prove that
\[\int\limits_{- a}^a f\left( x \right) dx = \int\limits_0^a \left\{ f\left( x \right) + f\left( - x \right) \right\} dx\]
\[\int\limits_0^1 \left( 3 x^2 + 5x \right) dx\]
\[\int\limits_0^2 x\left[ x \right] dx .\]
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Evaluate the following using properties of definite integral:
`int_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
