Advertisements
Advertisements
प्रश्न
\[\int\frac{\cos 4x - \cos 2x}{\sin 4x - \sin 2x} dx\]
योग
Advertisements
उत्तर
\[\int\left( \frac{\cos4x - \cos2x}{\sin4x - \sin2x} \right)dx\]
\[ = \int\frac{- 2\sin\left( \frac{4x + 2x}{2} \right)\sin\left( \frac{4x - 2x}{2} \right)}{2\cos\left( \frac{4x + 2x}{2} \right)\sin\left( \frac{4x - 2x}{2} \right)}dx \left[ \because \cos A - \cos B = - 2\sin \left( \frac{A + B}{2} \right)\sin \left( \frac{A - B}{2} \right) \text{and} \sin A - \sin B = 2\cos \left( \frac{A + B}{2} \right)\sin \left( \frac{A - B}{2} \right) \right]\]
\[ = - \int\frac{\sin 3x}{\cos 3x}dx\]
\[ = - \int\tan 3x dx\]
\[ = \frac{- \text{ln }\left| \sec 3x \right|}{3} + C\]
\[ = \frac{1}{3} \text{ln} \left( \left| \text{sec 3x} \right| \right)^{- 1} + C\]
\[ = \frac{1}{3} \text{ln }\left| \cos 3x \right| + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\left( \frac{m}{x} + \frac{x}{m} + m^x + x^m + mx \right) dx\]
\[\int\frac{1}{\sqrt{x}}\left( 1 + \frac{1}{x} \right) dx\]
\[\int\frac{x^{- 1/3} + \sqrt{x} + 2}{\sqrt[3]{x}} dx\]
\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]
\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]
\[\int \left( \tan x + \cot x \right)^2 dx\]
\[\int \left( a \tan x + b \cot x \right)^2 dx\]
\[\int\frac{1 + \cos 4x}{\cot x - \tan x} dx\]
\[\int\text{sin mx }\text{cos nx dx m }\neq n\]
\[\int\frac{1 - \cot x}{1 + \cot x} dx\]
` ∫ tan 2x tan 3x tan 5x dx `
\[\int \sin^5\text{ x }\text{cos x dx}\]
\[\int x^2 e^{x^3} \cos \left( e^{x^3} \right) dx\]
\[\int\frac{\left( \sin^{- 1} x \right)^3}{\sqrt{1 - x^2}} dx\]
\[\int \tan^3 \text{2x sec 2x dx}\]
\[\int \cot^6 x \text{ dx }\]
\[\int\frac{e^x}{\left( 1 + e^x \right)\left( 2 + e^x \right)} dx\]
\[\int\frac{1}{\sqrt{16 - 6x - x^2}} dx\]
\[\int\frac{1 - 3x}{3 x^2 + 4x + 2}\text{ dx}\]
\[\int\frac{x^2 + x + 1}{x^2 - x + 1} \text{ dx }\]
\[\int\frac{1}{4 \sin^2 x + 5 \cos^2 x} \text{ dx }\]
\[\int\frac{\sin 2x}{\sin^4 x + \cos^4 x} \text{ dx }\]
\[\int\frac{2 \tan x + 3}{3 \tan x + 4} \text{ dx }\]
` ∫ sin x log (\text{ cos x ) } dx `
\[\int \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) \text{ dx }\]
\[\int x^2 \tan^{- 1} x\text{ dx }\]
\[\int\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}} dx\]
\[\int e^x \frac{\left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[\int x\sqrt{x^2 + x} \text{ dx }\]
\[\int\frac{x}{\left( x^2 - a^2 \right) \left( x^2 - b^2 \right)} dx\]
\[\int\frac{dx}{\left( x^2 + 1 \right) \left( x^2 + 4 \right)}\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{x + 2}} \text{ dx }\]
\[\int\frac{x}{\left( x - 3 \right) \sqrt{x + 1}} dx\]
\[\int\frac{\sin x + \cos x}{\sqrt{\sin 2x}} \text{ dx}\]
\[\int\frac{1}{\sqrt{x^2 + a^2}} \text{ dx }\]
\[\int\frac{1}{\sqrt{3 - 2x - x^2}} \text{ dx}\]
\[\int\frac{1}{\left( \sin x - 2 \cos x \right) \left( 2 \sin x + \cos x \right)} \text{ dx }\]
\[\int\frac{1}{1 + 2 \cos x} \text{ dx }\]
\[\int \sin^{- 1} \sqrt{\frac{x}{a + x}} \text{ dx}\]
\[\int\frac{x^2}{\left( x - 1 \right)^3 \left( x + 1 \right)} \text{ dx}\]
