Advertisements
Advertisements
प्रश्न
\[\int x \sec^2 2x\ dx\]
योग
Advertisements
उत्तर
\[\int x_I \cdot \sec^2 2_{II}x\ dx \]
\[ = x\int \sec^2 2x\ dx - \int\left\{ \frac{d}{dx}\left( x \right)\int \sec^2 2x\ dx \right\}dx\]
\[ = \frac{x \tan 2x}{2} - \int1 \cdot \frac{\tan 2x}{2} dx\]
\[ = \frac{x \tan 2x}{2} - \frac{1}{2} \frac{\text{ ln } \left| \sec 2x \right|}{2} + C\]
\[ = \frac{x \tan 2x}{2} - \frac{1}{4} \text{ ln} \left| \sec 2x \right| + C\]
\[ = x\int \sec^2 2x\ dx - \int\left\{ \frac{d}{dx}\left( x \right)\int \sec^2 2x\ dx \right\}dx\]
\[ = \frac{x \tan 2x}{2} - \int1 \cdot \frac{\tan 2x}{2} dx\]
\[ = \frac{x \tan 2x}{2} - \frac{1}{2} \frac{\text{ ln } \left| \sec 2x \right|}{2} + C\]
\[ = \frac{x \tan 2x}{2} - \frac{1}{4} \text{ ln} \left| \sec 2x \right| + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\tan x}{\sec x + \tan x} dx\]
\[\int \cos^{- 1} \left( \sin x \right) dx\]
\[\int\frac{\text{sin} \left( x - \alpha \right)}{\text{sin }\left( x + \alpha \right)} dx\]
\[\int\sqrt{\frac{1 - \sin 2x}{1 + \sin 2x}} dx\]
` ∫ {sin 2x} /{a cos^2 x + b sin^2 x } ` dx
\[\int\frac{\left( x + 1 \right) e^x}{\cos^2 \left( x e^x \right)} dx\]
\[\int\left( \frac{x + 1}{x} \right) \left( x + \log x \right)^2 dx\]
\[\int\frac{\sin \left( \tan^{- 1} x \right)}{1 + x^2} dx\]
\[\int\frac{1}{\sqrt{x} + x} \text{ dx }\]
\[\int \sin^5 x \cos x \text{ dx }\]
\[\int\frac{x^4 + 1}{x^2 + 1} dx\]
\[\int\frac{x^2}{x^6 + a^6} dx\]
\[\int\frac{\sin 8x}{\sqrt{9 + \sin^4 4x}} dx\]
\[\int\frac{1 - 3x}{3 x^2 + 4x + 2}\text{ dx}\]
\[\int\frac{x^3 + x^2 + 2x + 1}{x^2 - x + 1}\text{ dx }\]
\[\int\frac{1}{4 \cos^2 x + 9 \sin^2 x}\text{ dx }\]
\[\int\frac{\log \left( \log x \right)}{x} dx\]
\[\int \sin^{- 1} \left( 3x - 4 x^3 \right) \text{ dx }\]
\[\int \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) \text{ dx }\]
\[\int\left( x + 1 \right) \text{ log x dx }\]
\[\int x^2 \tan^{- 1} x\text{ dx }\]
\[\int\left( e^\text{log x} + \sin x \right) \text{ cos x dx }\]
\[\int\left( 2x - 5 \right) \sqrt{2 + 3x - x^2} \text{ dx }\]
\[\int\frac{5x}{\left( x + 1 \right) \left( x^2 - 4 \right)} dx\]
\[\int\frac{2x - 3}{\left( x^2 - 1 \right) \left( 2x + 3 \right)} dx\]
\[\int\frac{\sin 2x}{\left( 1 + \sin x \right) \left( 2 + \sin x \right)} dx\]
\[\int\frac{1}{\left( x^2 + 1 \right) \sqrt{x}} \text{ dx }\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{x^2 + 1}} \text{ dx }\]
Write the anti-derivative of \[\left( 3\sqrt{x} + \frac{1}{\sqrt{x}} \right) .\]
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
\[\int\frac{1}{e^x + 1} \text{ dx }\]
\[\int\frac{1}{\sqrt{x^2 - a^2}} \text{ dx }\]
\[\int\frac{1}{1 + 2 \cos x} \text{ dx }\]
\[\int\frac{1}{\sin x + \sin 2x} \text{ dx }\]
\[\int \sec^6 x\ dx\]
\[\int \tan^5 x\ \sec^3 x\ dx\]
\[\int\frac{1}{x \sqrt{1 + x^n}} \text{ dx}\]
\[\int\frac{x^5}{\sqrt{1 + x^3}} \text{ dx }\]
\[\int \sin^{- 1} \sqrt{x}\ dx\]
