Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\text{ Let I }= \int\sqrt{2ax - x^2}\text{ dx}\]
\[ = \int\sqrt{a^2 + 2ax - x^2 - a^2}\text{ dx}\]
\[ = \int \sqrt{a^2 - \left( x^2 - 2ax + a^2 \right)}\text{ dx}\]
\[ = \int\sqrt{a^2 - \left( x - a \right)^2}\text{ dx}\]
\[ = \left( \frac{x - a}{2} \right) \sqrt{2ax - x^2} + \frac{a^2}{2} \sin^{- 1} \left( \frac{x - a}{a} \right) + C\]
APPEARS IN
संबंधित प्रश्न
If f' (x) = 8x3 − 2x, f(2) = 8, find f(x)
` = ∫ root (3){ cos^2 x} sin x dx `
` ∫ e^{m sin ^-1 x}/ \sqrt{1-x^2} ` dx
Evaluate the following integrals:
\[\int\frac{x}{\sqrt{8 + x - x^2}} dx\]
Evaluate the following integral:
\[\int\frac{x^3}{\sqrt{1 + x^2}}dx = a \left( 1 + x^2 \right)^\frac{3}{2} + b\sqrt{1 + x^2} + C\], then
\[\int\frac{1}{\sqrt{x} + \sqrt{x + 1}} \text{ dx }\]
\[ \int\left( 1 + x^2 \right) \ \cos 2x \ dx\]
