Advertisements
Advertisements
प्रश्न
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
योग
उत्तर
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
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
Evaluate the following definite integrals:
\[\int_0^\frac{\pi}{2} x^2 \sin\ x\ dx\]
\[\int\limits_0^1 x e^{x^2} dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_{- 1}^1 \left| 1 - x \right| dx\] is equal to
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx\] equals to
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
