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