Advertisements
Advertisements
प्रश्न
` ∫ sin x \sqrt (1-cos 2x) dx `
बेरीज
उत्तर
` ∫ sin x . \sqrt (1-cos 2x) dx `
` ∫ sin x \sqrt (2 sin^2 x ) dx ` `[∴ 1 - cos 2A = 2 sin^2 A]`
` = \sqrt2 ∫ sin^2 x dx `
\[ = \sqrt{2}\int\left( \frac{1 - \cos 2x}{2} \right)dx\]
\[ = \frac{1}{\sqrt{2}}\int\left( 1 - \cos 2x \right)dx\]
\[ = \frac{1}{\sqrt{2}}\left[ x - \frac{\sin 2x}{2} \right] + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{x^{- 1/3} + \sqrt{x} + 2}{\sqrt[3]{x}} dx\]
\[\int\frac{\sin^2 x}{1 + \cos x} \text{dx} \]
\[\int \cot^{- 1} \left( \frac{\sin 2x}{1 - \cos 2x} \right) dx\]
\[\int\frac{1}{\left( 7x - 5 \right)^3} + \frac{1}{\sqrt{5x - 4}} dx\]
\[\int\frac{x}{\sqrt{x + a} - \sqrt{x + b}}dx\]
Integrate the following integrals:
\[\int\text { sin x cos 2x sin 3x dx}\]
\[\int\frac{\log\left( 1 + \frac{1}{x} \right)}{x \left( 1 + x \right)} dx\]
\[\int \sin^5\text{ x }\text{cos x dx}\]
` ∫ sec^6 x tan x dx `
\[\int \sin^7 x \text{ dx }\]
\[\int\frac{\sin x}{\sqrt{4 \cos^2 x - 1}} dx\]
\[\int\frac{\cos x - \sin x}{\sqrt{8 - \sin2x}}dx\]
\[\int\frac{x^2}{x^2 + 6x + 12} \text{ dx }\]
\[\int\frac{x + 1}{\sqrt{x^2 + 1}} dx\]
\[\int\sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]
\[\int\frac{\sin 2x}{\sin^4 x + \cos^4 x} \text{ dx }\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int\frac{1}{1 - \sin x + \cos x} \text{ dx }\]
`int"x"^"n"."log" "x" "dx"`
\[\int\cos\sqrt{x}\ dx\]
\[\int \sec^{- 1} \sqrt{x}\ dx\]
\[\int \cos^{- 1} \left( 4 x^3 - 3x \right) \text{ dx }\]
\[\int \sin^3 \sqrt{x}\ dx\]
\[\int\sqrt{2x - x^2} \text{ dx}\]
\[\int\left( 2x - 5 \right) \sqrt{2 + 3x - x^2} \text{ dx }\]
\[\int\frac{2x - 3}{\left( x^2 - 1 \right) \left( 2x + 3 \right)} dx\]
\[\int\frac{x^2}{\left( x - 1 \right) \left( x + 1 \right)^2} dx\]
\[\int\frac{x^2 + 1}{x^4 + 7 x^2 + 1} 2 \text{ dx }\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{2x + 3}} \text{ dx }\]
\[\int\frac{x}{\left( x^2 + 2x + 2 \right) \sqrt{x + 1}} \text{ dx}\]
` \int \text{ x} \text{ sec x}^2 \text{ dx is equal to }`
If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then
\[\int\frac{1 - x^4}{1 - x} \text{ dx }\]
\[\int\frac{1}{a + b \tan x} \text{ dx }\]
\[\int \log_{10} x\ dx\]
\[\int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]
\[\int \left( \sin^{- 1} x \right)^3 dx\]
\[\int\frac{x^2 - 2}{x^5 - x} \text{ dx}\]
