Advertisements
Advertisements
प्रश्न
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
योग
Advertisements
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
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^1 \frac{1}{2 x^2 + x + 1} dx\]
\[\int\limits_0^4 \frac{1}{\sqrt{4x - x^2}} dx\]
\[\int_0^1 x\log\left( 1 + 2x \right)dx\]
\[\int_{- 2}^2 x e^\left| x \right| dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} dx\]
\[\int\limits_0^1 \left( 3 x^2 + 5x \right) dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \cos^2 x\ dx .\]
\[\int\limits_{- 1}^1 \left| 1 - x \right| dx\] is equal to
\[\int\limits_{- 1}^1 e^{2x} dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
