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
\[\int\limits_0^1 \left( x e^{2x} + \sin\frac{\ pix}{2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{1}{5 \cos x + 3 \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x \cos x}{1 + \sin^4 x} dx\]
\[\int_0^\frac{\pi}{2} \frac{\cos^2 x}{1 + 3 \sin^2 x}dx\]
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
\[\int\limits_a^b x\ dx\]
\[\int\limits_1^\sqrt{3} \frac{1}{1 + x^2} dx\] is equal to ______.
\[\int\limits_0^{\pi/2} x \sin x\ dx\] is equal to
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
