Advertisements
Advertisements
प्रश्न
\[\int\sqrt{2x - x^2} \text{ dx}\]
बेरीज
उत्तर
\[I = \int\sqrt{2x - x^2}\text{ dx}\]
\[ = \int\sqrt{x\left( 2 - x \right)}\text{ dx}\]
\[ = \int\sqrt{x\left( 2 - x \right)}\text{ dx}\]
Let
\[x = 1 + \ sin\ u\]
\[or, dx = \cos\ u\ du\]
\[ \Rightarrow I = \int\sqrt{\left( 1 + \sin u \right)\left( 1 - \sin u \right)}\ cos\ u\ du\]
\[ \Rightarrow I = \int \cos^2 u\ du\]
\[ \Rightarrow I = \frac{1}{2}\int\left( \cos2u + 1 \right)du\]
\[ \Rightarrow I = \int\sqrt{\left( 1 + \sin u \right)\left( 1 - \sin u \right)}\ cos\ u\ du\]
\[ \Rightarrow I = \int \cos^2 u\ du\]
\[ \Rightarrow I = \frac{1}{2}\int\left( \cos2u + 1 \right)du\]
\[\Rightarrow I = \frac{1}{2}\left( \frac{1}{2}\sin 2u + u \right) + c\]
\[ \Rightarrow I = \frac{1}{2}\left( \sin u \cos u + u \right) + c\]
\[ \Rightarrow I = \frac{1}{2}\left( \sin u \sqrt{1 - \sin^2 u} + u \right) + c\]
\[ \therefore I = \frac{1}{2}\left( x - 1 \right)\sqrt{2x - x^2} + \frac{1}{2} \sin^{- 1} \left( x - 1 \right) + c \left[ \because u = \sin^{- 1} \left( x - 1 \right) \right]\]
\[ \Rightarrow I = \frac{1}{2}\left( \sin u \cos u + u \right) + c\]
\[ \Rightarrow I = \frac{1}{2}\left( \sin u \sqrt{1 - \sin^2 u} + u \right) + c\]
\[ \therefore I = \frac{1}{2}\left( x - 1 \right)\sqrt{2x - x^2} + \frac{1}{2} \sin^{- 1} \left( x - 1 \right) + c \left[ \because u = \sin^{- 1} \left( x - 1 \right) \right]\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\left( 2 - 3x \right) \left( 3 + 2x \right) \left( 1 - 2x \right) dx\]
\[\int\frac{\left( 1 + x \right)^3}{\sqrt{x}} dx\]
\[\int\frac{1 + \cos x}{1 - \cos x} dx\]
\[\int\frac{\sin 2x}{\sin 5x \sin 3x} dx\]
` = ∫ root (3){ cos^2 x} sin x dx `
\[\int x^3 \sin x^4 dx\]
` ∫ x {tan^{- 1} x^2}/{1 + x^4} dx`
\[\int \tan^3 \text{2x sec 2x dx}\]
\[\int\frac{1}{\sqrt{x} + \sqrt[4]{x}}dx\]
` ∫ tan^5 x dx `
\[\int\frac{x^2}{x^6 + a^6} dx\]
\[\int\frac{1}{\sqrt{7 - 3x - 2 x^2}} dx\]
\[\int\frac{\cos x}{\sqrt{4 - \sin^2 x}} dx\]
` ∫ \sqrt{"cosec x"- 1} dx `
\[\int\frac{a x^3 + bx}{x^4 + c^2} dx\]
\[\int\frac{1}{5 + 7 \cos x + \sin x} dx\]
\[\int\frac{4 \sin x + 5 \cos x}{5 \sin x + 4 \cos x} \text{ dx }\]
\[\int x \cos x\ dx\]
`int"x"^"n"."log" "x" "dx"`
\[\int\frac{\left( x \tan^{- 1} x \right)}{\left( 1 + x^2 \right)^{3/2}} \text{ dx }\]
\[\int\sqrt{3 - x^2} \text{ dx}\]
\[\int\sqrt{x^2 - 2x} \text{ dx}\]
\[\int\frac{2x + 1}{\left( x + 1 \right) \left( x - 2 \right)} dx\]
\[\int\frac{x^2}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{18}{\left( x + 2 \right) \left( x^2 + 4 \right)} dx\]
\[\int\frac{dx}{\left( x^2 + 1 \right) \left( x^2 + 4 \right)}\]
\[\int\frac{x + 1}{\left( x - 1 \right) \sqrt{x + 2}} \text{ dx }\]
\[\int x^{\sin x} \left( \frac{\sin x}{x} + \cos x . \log x \right) dx\] is equal to
\[\int\frac{e^x \left( 1 + x \right)}{\cos^2 \left( x e^x \right)} dx =\]
The primitive of the function \[f\left( x \right) = \left( 1 - \frac{1}{x^2} \right) a^{x + \frac{1}{x}} , a > 0\text{ is}\]
\[\int\frac{x^3}{\sqrt{1 + x^2}}dx = a \left( 1 + x^2 \right)^\frac{3}{2} + b\sqrt{1 + x^2} + C\], then
\[\int \text{cosec}^2 x \text{ cos}^2 \text{ 2x dx} \]
\[\int\frac{\sin 2x}{a^2 + b^2 \sin^2 x}\]
\[\int x \sin^5 x^2 \cos x^2 dx\]
\[\int\frac{\sin^5 x}{\cos^4 x} \text{ dx }\]
\[\int\frac{\sin^6 x}{\cos x} \text{ dx }\]
\[\int\sqrt{a^2 + x^2} \text{ dx }\]
\[\int \sin^{- 1} \sqrt{x}\ dx\]
\[\int\frac{x^2 + 1}{x^2 - 5x + 6} \text{ dx }\]
