Advertisements
Advertisements
प्रश्न
` ∫ cos 3x cos 4x` dx
बेरीज
Advertisements
उत्तर
\[\int\text{cos 4x }\text{cos 3x dx}\]
` = 1/2 ∫ 2 cos 4x cos 3x dx `
\[ = \frac{1}{2}\int\left[ \text{cos} \left( 4x + 3x \right) + \text{cos }\left( 4x - 3x \right) \right]dx \left[ \therefore \text{2 }\text{cos A }\text{cos B} = \text{cos} \left( A + B \right) + \text{cos }\left( A - B \right) \right]\]
\[ = \frac{1}{2}\int\left( \text{cos} \left( 7x \right) + \text{cos x} \right) dx\]
\[ = \frac{1}{2}\left[ \frac{\sin 7x}{7} + \sin x \right] + C\]
\[ = \frac{1}{14}\sin 7x + \frac{1}{2}\sin x + C\]
\[ = \frac{1}{2}\int\left[ \text{cos} \left( 4x + 3x \right) + \text{cos }\left( 4x - 3x \right) \right]dx \left[ \therefore \text{2 }\text{cos A }\text{cos B} = \text{cos} \left( A + B \right) + \text{cos }\left( A - B \right) \right]\]
\[ = \frac{1}{2}\int\left( \text{cos} \left( 7x \right) + \text{cos x} \right) dx\]
\[ = \frac{1}{2}\left[ \frac{\sin 7x}{7} + \sin x \right] + C\]
\[ = \frac{1}{14}\sin 7x + \frac{1}{2}\sin x + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]
\[\int\frac{\tan x}{\sec x + \tan x} dx\]
` ∫ {cosec x} / {"cosec x "- cot x} ` dx
Write the primitive or anti-derivative of
\[f\left( x \right) = \sqrt{x} + \frac{1}{\sqrt{x}} .\]
\[\int\frac{1}{\left( 7x - 5 \right)^3} + \frac{1}{\sqrt{5x - 4}} dx\]
\[\int \left( e^x + 1 \right)^2 e^x dx\]
\[\int\frac{x + 1}{\sqrt{2x + 3}} dx\]
\[\int\frac{x}{\sqrt{x + a} - \sqrt{x + b}}dx\]
\[\int\frac{1 - \sin x}{x + \cos x} dx\]
\[\int\frac{\sin 2x}{\sin \left( x - \frac{\pi}{6} \right) \sin \left( x + \frac{\pi}{6} \right)} dx\]
\[\int\frac{\cos^5 x}{\sin x} dx\]
\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} dx\]
` ∫ tan^5 x sec ^4 x dx `
\[\int \sin^5 x \text{ dx }\]
\[\int\frac{1}{\sqrt{1 + 4 x^2}} dx\]
\[\int\frac{1}{4 x^2 + 12x + 5} dx\]
` ∫ { x^2 dx}/{x^6 - a^6} dx `
\[\int\frac{1}{\sqrt{\left( x - \alpha \right)\left( \beta - x \right)}} dx, \left( \beta > \alpha \right)\]
\[\int\frac{x}{x^2 + 3x + 2} dx\]
\[\int\frac{2x + 1}{\sqrt{x^2 + 4x + 3}} \text{ dx }\]
\[\int\frac{1}{3 + 2 \cos^2 x} \text{ dx }\]
\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]
\[\int\frac{1}{4 + 3 \tan x} dx\]
\[\int x^3 \text{ log x dx }\]
\[\int x^3 \cos x^2 dx\]
\[\int x \sin x \cos x\ dx\]
\[\int x^2 \tan^{- 1} x\text{ dx }\]
\[\int x^3 \tan^{- 1}\text{ x dx }\]
\[\int\left( \tan^{- 1} x^2 \right) x\ dx\]
\[\int e^x \left( \frac{x - 1}{2 x^2} \right) dx\]
\[\int e^x \left( \log x + \frac{1}{x} \right) dx\]
\[\int\sqrt{2ax - x^2} \text{ dx}\]
\[\int\frac{x^2}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{x^2 + 1}} \text{ dx }\]
Write the anti-derivative of \[\left( 3\sqrt{x} + \frac{1}{\sqrt{x}} \right) .\]
\[\int\frac{1}{7 + 5 \cos x} dx =\]
\[\int\sqrt{\text{ cosec x} - 1} \text{ dx }\]
\[\int\frac{\sin^5 x}{\cos^4 x} \text{ dx }\]
\[\int\sqrt{3 x^2 + 4x + 1}\text{ dx }\]
